acids and bases note sap

8/10/2019 Acids and Bases Note Sap

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IV. Acids and Bases

Properties of Acids and Bases

Acids Bases pH7.0taste sour taste bitterreact with metals to produce hydrogen gas feel slippery

pH paper turns red/pink pH paper turns blue/green

phenolphthalein colourless phenolphthalein pink bromothymol blue yellow bromothymol blue blue

pH and pOHpH (power of hydrogen) is a measure of the acidity of a solution: the lower the pH, the more acidic the solution is and thehigher the pH, the more basic the solution is. A solution is acidic when pH[OH ]. A solution is basic when pH>7and [H +]


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pH, pOH, and Significant Figures For pH and pOH, only the digits after the decimal place are significant.(ie, pH= 10.20 has two significant digits since there are two numbers after the decimal place; pOH = 1.384 has three significantdigits since there are three numbers after the decimal place)

ex. A solution has an [H +] of 0.040 M, calculate the pH. pH = log [H +] = log(0.040) = 1.40

ex. A solution has an [OH ] of 6.5x10 5 M, calculate the pOH. pOH = log[OH ] = log(6.5x10 5) = 4.19

ex. acid rain has a pH of 4.80, calculate the [H +]. [H +] = 10 pH = 10 4.80 = 1.5x10 5 M

ex. ammonia has a pOH of 3.78, calculate the [OH ] [OH ] = 10 pOH = 10 3.78 = 1.7x10 4 M

ex. calculate the [H +], pH, pOH, and [OH ] of 0.0020 M HCl

HCl H + + Cl

therefore, [H +] = 0.00200 M pH = log[H +]= log(0.0020)= 2.70

pH + pOH = 14.002.20 + pOH = 14.00

pOH = 11.30

[H +][OH ] = 1.0x10 (0.0020) [OH ] = 1.0x10 14 [OH ] = 5.0x10 12 M

ex. calculate the [OH ], pOH, pH, and [H +] of 0.010 M Sr(OH) 2

Sr(OH) 2 Sr 2+ + 2OH

therefore, [OH ] = 2(0.010 M)= 0.020 M

pOH = log[OH ]= log (0.020) = 1.70

pH + pOH = 14.00 pH + 1.70 = 14.00 pH = 12.30

[H +][OH ] = 1.0x10 14 [H +][0.020] = 1.0x10 14 [H +] = 5.0x10 13 M

The Ahhrenius Concept does not adequately describe some compounds that exhibit acidic or basic properties, so a more generaldefinition is required.

Bronsted Lowry ModelAcid: a proton (H +) donor Base: a proton (H +) acceptorConjugate acid: has one more proton Conjugate base: has one few proton

Acid/Base EquilibriaA weak acid or base is one that does not completely dissociate in water; instead it forms an equilibrium with water.

Acids: HA (aq) + H 2O (l) H 3O+

(aq) + A

(aq)acid base conjugate conjugateacid base

note: The proton is donated by the acid (HA) and accepted by the base (H 2O). The conjugate acid of water is the hydronium ion(H 3O

+) which has one additional proton and the conjugate base of HA is A which has one fewer proton..

The equilibrium expression for this reaction can be written as follows:

3[ ][ ][ ]a

H O A K

HA

Where K a is the acid dissociation constant.

The acid dissociation constant can also be expressed as pK a where pK a = log(K a) or 10 a pK a K

Bases: B (aq) + H 2O (l) OH

(aq) + BH+

(aq) base acid conjugate conjugate

base acid

note: The proton is accepted by the base (B) and donated by the acid (H 2O). The conjugate base of water is the hydroxide ion(OH ) which has one fewer proton and the conjugate acid of B is BH + which has one additional proton.

The equilibrium expression for this reaction can be written as follows:

[ ][ ][ ]b

OH BH K

B

Where K b is the base dissociation constant.

The base dissociation constant can also be expressed as pK b where pK b = log(K b) or 10 b pK b K


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K a/pK a values for Common Acids. The higher the K a/pK a value, the stronger the acid.Acid Formula K a pK a iodic acid HIO 3 0.17 0.77oxalic acid H 2C2O4 5.9x10

2 1.23sulphurous acid H 2SO 3 1.5x10

2 1.82hydrogen sulphate ion HSO 4

1.2x10 2 1.92chlorous acid HClO 2 1.2x10

2 1.92 phosphoric acid H 3PO 4 7.5x10

3 2.12

hydrofluoric acid HF 7.2x10 4

3.14nitrous acid HNO 2 4.0x10 4 3.40

hydrogen oxalate ion HC 2O4 6.1x10 5 4.21

acetic acid HC 2H3O2 (orCH 3COOH)1.8x10 5 4.74

carbonic acid H 2CO 3 4.3x10 7 6.37

hydrogen sulphite ion HSO 3 1.0x10 7 7.00

hydrosulphuric acid H 2S 1.0x10 7 7.00

dihydrogen phosphateion H2PO 4

6.2x10 8 7.21hypochlorous acid HOCl 3.5x10 8 7.46hypobromous acid HOBr 2.0x10 9 8.70

hydrocyanic acid HCN 6.2x10 10

9.21 boric acid H 3BO 3 5.8x10

10 9.24ammonium ion NH 4

+ 5.6x10 10 9.25hydrogen carbonate ion HCO 3

5.6x10 11 10.25hydrogen phosphate ion HPO 4

2 4.8x10 13 12.32hydrogen sulphide ion HS 1.3x10 13 12.89

K b/pK b values for Common Bases. The higher the K b//pK b value, the stronger the base.Base Formula Conjugate Acid K b pK b diethylamine C 4H10 NH C 4H10 NH 2

+ 1.3x10 3 2.89ethylamine C 2H5 NH 2 C2H5 NH 3

+ 5.6x10 4 3.25methylamine CH 3 NH 2 CH 3 NH 3

+ 4.4x10 4 3.36triethylamine C 6H15 N C 6H15 NH + 4.0x10 4 3.40ammonia NH 3 NH 4

+ 1.8x10 5 4.74hydrazine H 2 NNH 2 (or N 2H4) H 2 NNH 3

+ (or N 2H5+) 3.0x10 6 5.52

hydroxylamine HONH 2 (or ONH 3) HONH 3+ (or ONH 4

+) 1.1x10 8 7.96 pyridine C 5H5 N C 5H5 NH

+ 1.7x10 9 8.77aniline C 6H5 NH 2 C6H5 NH 3

+ 3.8x10 10 9.42

ex. HF (aq) + H 2O (l) H 3O+ (aq) + F (aq)

acid base conjugate conjugateacid base

3[ ][ ][ ]a

H O F K

HF

ex. NH 3 (aq) + H 2O (l) OH (aq) + NH 4

+ (aq) base acid conjugate conjugate

base acid4

3[ ][ ][ ]b OH NH K NH

ex. Write a K a or K b expression for the following acid-base equilibria. Identify the conjugate acid base pairs.(As a general guideline: For acids, the hydrogen is typically taken away from the beginning of the formula; for bases, thehydrogen is typically added at the end of the formula)

(1) HC 2H3O2 (aq) + H 2O (l) H 3O+ (aq) + C 2H3O2

(aq)

(2) CH 3 NH 2 (aq) + H 2O (l) OH (aq) + CH 3 NH 3

+ (aq)


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ex. Complete the following acid base equilibria. Identify the conjugate acid base pairs and write a K a or K b expression.

(1) HOBr (aq) + H 2O (l)

(2) C 6H5 NH 2 (aq) + H 2O (l)

Hydrated Metal IonsCertain metal ions (particularly transition metal ions and aluminum) when in solution will form hydrated complexes that may actas acid.

ex. When Al 3+ is in solution, it forms [Al(H 2O) 6]3+, the acid equilibrium for this complex is as follows: (K a = 1.4x10

5)[Al(H 2O) 6]

3+ (aq) + H 2O (l) H 3O+ (aq) + [Al(H 2O) 5OH]

2+ (aq)

ex. When Fe 3+ is in solution, it forms [Fe(H 2O) 6]3+. Complete the acid equilibrium for this complex. (K a = 6.0x10

3)[Fe(H 2O)6]

3+ (aq) + H 2O (l)

Calculations for Weak Acids and Bases

The equilibrium concentration of the hydronium ion or hydroxide ion must be determined using an ICE table.

ex. Determine the pH and pOH of 1.0 M solution of HF (K a = 7.2x10 4).

HF (aq) + H 2O (l) H 3O+ (aq) + F (aq)

I 1.0 0 0

C x +x +x

E 1.0 x x x

assume x is small 1.0 x 1.0

3[ ][ ][ ]a

H O F K

HF

247.2 10

1.0 x

x

x = 0.027 M = [H 3O+]

100%

0.027100% 2.7%

1.0

changeinconcentration percentchangein concentration x

initial concentration

x

(therefore the approximation is valid)

pH = log[H +]= log[H 3O+]

= log(0.027)= 1.57(H 3O

+ is equivalent to H + for pH calculations)

pH + pOH = 14.001.57 + pOH = 14.00

pOH = 12.43

ex. Determine the pOH and pH of a 1.0 M solution of NH 3 (K b = 1.8x10 5).

NH 3 (aq) + H 2O (l) OH (aq) + NH 4

+ (aq)

I 1.0 0 0

C x +x +xE 1.0 x x x


4

3

[ ][ ][ ]b

OH NH K

NH

251.8 10

1.0 x

x

x = 0.0042 M = [OH ]

100%

0.0042100% 0.42%

1.0

changeinconcentration percentchangein concentration x

initial concentration

x

(therefore the approximation is valid)

pOH = log[OH ]= log(0.0042)

= 2.38

pH + pOH = 14.00 pH + 2.38 = 14.00 pH = 11.62


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ex. The pH of a 0.650 M solution of citric acid (H 3C6H5O7) is 1.63. Determine K a and pK a.

ex. The pOH of a 0.020 M solution of toluidine (C 7H7 NH 2) is 5.35. Determine K b and pK b.

Percent Dissociation (also called percent ionization) The percent dissociation of a compound is a measure of the degree of dissociation.For strong acids and bases, the percent dissociation is 100% (since the compound completely dissociates into its constituentions). For weak acids and bases, the percent dissociation can be calculated using an ICE table and according to the followingequation:

100%amount dissociated

Percent dissociation xinitial concentration

amount dissociated = x

ex. Calculate the percent dissociation for a 0.22 M solution of C 2H5 NH 2 (K b = 5.6x10 4).

ex. A 0.1000 M solution of lactic acid (HC 3H5O3) is 3.7% dissociated at equilibrium. Calculate the K a value for lactic acid.


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Comparing the pH of a strong acid and a weak acid with the same concentration A solution of a strong acid will have a much lower pH that a solution of a weak acid with the same initial concentration.

ex. (a) What is the pH of a solution with [HCl] = 0.10 M(b) What is the pH of a solution with [HC 2H3O2] = 0.10 M

(a) For 0.10 M HCl, the HCl completely dissociates.HCl H + + Cl

therefore, [H +] = 0.10 M pH = log[H +]= log(0.10)= 1.00

(b) Since HC 2H3O2 does not completely dissociate, the pH must be found using an ICE table.

HC 2H3O2 (aq) + H2O (l) H3O+ (aq) + C 2H3O2

(aq)

I 0.10 0 0

C x +x +x

E 0.10 x x x


3 2 3 2

2 3 2

[ ][ ][ ]a

H O C H O K

HC H O

251.8 10

0.10 x

x

x = 0.0013 M = [H 3O+]

pH = log[H 3O+]

= log(0.0 013)= 2.87

The pH of a solution of the strong acid HCl with a concentration of 0.10 M (1.00) is much lower/more acidic than the pH of asolution of the weak acid HC 2H3O2 with an initial concentration of 0.10 M (2.87).

Comparing the concentration of a strong acid and a weak acid with the same pHA solution of a strong acid will have a much lower concentration that a solution of a weak acid with the same pH.

ex. (a) What is the concentration of a solution of HCl with a pH of 2.00?(b) What is the concentration of a solution of HNO 2 with a pH of 2.00?

pH = 2.00, therefore [H +] (or [H 3O+]) = 10 pH = 10 2.00 = 0.010 M

(a) If [H +] = 0.010 M, since HCl completely dissociates, [HCl] = 0.010 M

(b) Since HNO 2 does not completely dissociate, the concentration of HNO 2 must be found using an ICE table. The equilibriumconcentration of HNO 2 can be defined as "x". The equilibrium concentration of [H 3O

+] = 0.010 M and will be equal to theequilibrium [NO 2

]. The initial concentration of HNO 2 will therefore be x + 0.010.

HNO 2 (aq) + H2O (l) H3O+ (aq) + NO 2

(aq)

I x + 0.010 0 0

C 0.010 +0.010 +0.010

E x 0.010 0.010

3 2

2

[ ][ ][ ]a

H O NO K

HNO

24 (0.010)

4.0 10 x x x = 0.25 M

therefore, the initial[HNO 2] = 0.25 + 0.010= 0.26 M

The initial concentration of a solution of the strong acid HCl with a pH of 2.00 (0.010 M) is much lower that the initialconcentration of a solution of the weak acid HNO 2 with a pH of 2.00 (0.26 M).


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Amphoteric CompoundsAmphoteric : A compound that can act as either an acid or a base.

ex. Water has been shown to act like both an acid (when in solution with a weak base) and as a base (when in solution with aweak acid). Pure water will also undergo autoionization (self-ionization) such that some water molecules will act as acids andsome water molecules will act as bases according to the following reaction:

H2O (l) + H 2O (l) H 3O+ (aq) + OH (aq)

note: The autoionization of water produces both hydronium ions hydroxide ions at equal concentrations ([H+

] = [OH

]) so theresulting solution is neutral.

The equilibrium expression for this reaction is as follows:

3[ ][ ]w K H O OH

, where K w is called the dissociation constant of water. K w = 1.0x10 14 (at 25 C) or pK w = 14

Neutral Solutions at Non-Standard TemperaturesIn order for a solution to be neutral, the requirement is that [H 3O

+] = [OH ]. (Not that pH = 7)At 25 C, since K w = 1.0x10

14, [H 3O+] = [OH ] = 1.0x10 7 which results in a pH (and pOH) of 7.0.

The autoionization of water is an endothermic process such that at higher temperatures the equilibrium shifts right and the valueof K w increases. Since the dissociation constant of water (K w) increases with temperature, [H

+] and [OH ] in a neutral solutionwill increase with temperature and as a result, the pH of a neutral solution will decrease with temperature.

ex. At 40 C, K w = 2.9x10 14

At this temperature, [H 3O

+] = [OH ] = 1.7x10 7 M, so neutral, pure water will have a pH (and pOH) of 6.78

ex. At 10 C, K w = 2.9x10 15

At this temperature, [H 3O+] = [OH ] = 5.4x10 8 M, so neutral, pure water will have a pH (and pOH) of 7.27.

(note: pH + pOH = 14 only applies for solutions at 25 C, the more general equation is that pH + pOH = pK w)

Relationship between K a and K b For a weak acid, if the K a value is known, then the K b value of its conjugate base can be calculated. For a weak base, if the K b value is known, then the K a value for its conjugate acid can be calculated. This relationship is described according to thefollowing equation:

a b w K K K K w = 1.0x10 14 (at 25 C), K a is the acid dissociation constant, and K b is the base dissociation constantThis equation can also be written as pK a + pK b = 14.00

ex. Hydrofluoric acid, HF has K a = 7.2x10 4

HF (aq) + H 2O (l) H 3O+ (aq) + F (aq)

acid conjugate base3[ ][ ][ ]a

H O F K

HF

= 7.2x10 4

Calculate K b for F . Give the equilibrium reaction and K b expression.

Since F is the conjugate base of HF,14

114

1.0 101.4 10

7.2 10

wb

a

K x K x

K x

F (aq) + H 2O (l) OH (aq) + HF (aq)

[ ][ ][ ]b

OH HF K

F = 1.4x10 11

Note: 3 3[ ][ ] [ ][ ]

[ ][ ][ ] [ ]a b w

H O F OH HF K xK x H O OH K

HF F


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ex. Pyridine, C 5H5 N has K b = 1.7x10 9

C5H5 N (aq) + H 2O (l) OH (aq) + C 5H5 NH

+ (aq) base conjugate acid

5 5

5 5

[ ][ ][ ]b

OH C H NH K

C H N

=1.7x10 9

Calculate K a for C 5H5 NH+. Give the equilibrium reaction and K a expression.

Since C 5H5 NH+ is the conjugate acid of C 5H5 N,

149

9

1.0 105.9 10

1.7 10w

ab

K x K x

K x

C5H5 NH+ (aq) + H 2O (l) H 3O

+ (aq) + C 5H5 N (aq) 3 5 5

5 5

[ ][ ][ ]a

H O C H N K C H NH

= 5.9x10 9

Note: 5 5 3 5 5 35 5 5 5

[ ][ ] [ ][ ][ ][ ]

[ ] [ ]b a wOH C H NH H O C H N

K xK x H O OH K C H N C H NH

Strength of a Conjugate Acid or BaseThe stronger the acid, the weaker its conjugate base and the stronger the base, the weaker its conjugate acid.

ex. (a) Which of the following substances is a stronger acid, HF or HCN?

(b)Which of the following substances is a stronger base, F

or CN

?

(a) HF has K a = 7.2x10 4 and HCN has K a = 6.2x10

10. Since K a for HF is greater than K a for HCN, HF is a stronger acid thanHCN.

(b) F has K b = 1.0x10 14/7.2x10 4 = 1.4x10 11 and CN has K b = 1.0x10

14/6.2x10 10 = 1.6x10 5. Since K b for CN is greater than

K b for F , CN is a stronger base than F .

HF is a stronger acid than HCN, so CN is a stronger base that F

ex. (a) Which of the following substances is a stronger base, NH 3 or C 5H5 N?(b)Which of the following substances is a stronger acid, NH 4

+ or C 5H5 NH+?

(a) NH 3 has K b = 1.8x10 5 and C 5H5 N has K b = 1.7x10

9. Since K b for NH 3 is greater than K b for C 5H5 N, NH 3 is a stronger basethan C 5H5 N.

(b) NH 4+ has K a = 5.6x10

10 (given on K a table) and C 5H5 NH+ has K a = 1.0x10

14/1.7x10 9 = 5.9x10 9. Since K a for C 5H5 NH+ is

greater than K a for NH 4+, C 5H5 NH

+ is a stronger acid than NH 4+.

NH 3 is stronger base than C 5H5 N, so C 5H5 NH+ is a stronger acid than NH 4

+.

Amphoteric Compounds and K a/K b In addition to water, there are many other amphoteric substances including the hydrogen carbonate ion (HCO 3

), the hydrogensulphate ion (HSO 4

), the hydrogen sulphite ion (HSO 3 ), the dihydrogen phosphate ion (H 2PO 4

), the hydrogen phosphate ion(HPO 4

2 ), the hydrogen sulphide ion (HS ) etc.

Comparing Equilibria for Amphoteric Compoundsex. the hydrogen carbonate ion, HCO 3

can act as an acid (donating a proton) or a base (accepting a proton)

reaction as an acid: HCO 3 (aq) + H 2O (l) H 3O

+ (aq) + CO 32 (aq) K a = 5.6x10

11 (from K a chart)

reaction as a base: HCO 3 (aq) + H 2O (l) OH

(aq) + H 2CO 3 (aq)14

87

1.0 102.3 10

4.3 10w

ba

K x K x

K x

(Note: K a for H 2CO 3 is used sinceH2CO 3 is the conjugate acid HCO 3

)

Since K b>K a, a solution containing hydrogen carbonate would have a basic pH (since [OH ] will be greater than [H 3O

+])


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Acid Base ReactionsWeak acids can react with weak bases and result in an acid/base equilibrium.

Consider the following reactions:

(1) NH 3 (aq) + H 2S (aq) HS (aq) + NH 4

+ (aq) base acid conjugate conjugate

base acid

Calculating K for an acidbase reaction: Consider the equilibria for the compounds acting as acids in the reaction:H2S (aq) + H 2O (l) H 3O

+ (aq) + HS (aq) K a = 1.0x10 7

NH 4+ (aq) + H 2O (l) H 3O

+ (aq) + NH 3 (aq) K a = 5.6x10 10

Note, if the second equilibrium is reversed (and the reciprocal of the equilibrium constant is taken), the two equilibria can add upto give the overall equilibrium (and the K for the equilibrium can be obtained by multiplying K for each of the reactions)

H2S (aq) + H 2O (l) H 3O+ (aq) + HS (aq) K a = 1.0x10

7 H3O

+ (aq) + NH 3 (aq) NH 4+ (aq) + H 2O (l) K = 1/5.6x10

10 = 1.8x10 9 __________________________________________________________

NH 3 (aq) + H 2S (aq) HS (aq) + NH 4

+ (aq) K =1.0x10 7 x 1.8x10 9 = 1.8x10 2 (note: H 3O

+ (aq) and H 2O (l) are cancelled when the equilibria are added)An important result is that K>1. This is significant since H 2S is a stronger acid than NH 4

+, therefore the equilibrium position isdriven to the right, favouring the products.

This same result could be derived by considering the equilibria for the compounds that act as bases in the reaction: NH 3 (aq) + H 2O (l) OH

(aq) + NH 4+ (aq) K b = 1.8x10

5 HS (aq) + H 2O (l) OH

(aq) + H 2S (aq) K b = 1.0x10 14/1.0x10 7 = 1.0x10 7

Note, if the second equilibrium is reversed (and the reciprocal of the equilibrium constant is taken), the two equilibria can add upto give the overall equilibrium (and the K for the equilibrium can be obtained by multiplying K for each of the reactions)

NH 3 (aq) + H 2O (l) OH (aq) + NH 4

+ (aq) K b = 1.8x10 5

OH (aq) + H 2S (aq) HS (aq) + H 2O (l) K = 1/1.0x10 7 = 1.0x10 7 ______________________________________________________

NH 3 (aq) + H 2S (aq) HS (aq) + NH 4

+ (aq) K =1.8x10 5 x 1.0x10 7 = 1.8x10 2 (as determined above)

(note: OH (aq) and H 2O (l) are cancelled when the equilibria are added)

The result that K>1 also correspond to the fact that NH 3 is a stronger base than HS , therefore the equilibrium position is driven

to the right, favouring the products.

(2) HOCl (aq) + C 6H5 NH 2 (aq) OCl (aq) + C 6H5 NH 3

+ (aq)

Consider the equilibria for the compounds acting as acids in the reaction:HOCl (aq) + H 2O (l) H 3O+ (aq) + OCl (aq) K a =3.5x10 8 C6H5 NH 3

+ (aq) + H 2O (l) H 3O+ (aq) + C 6H5 NH 2 (aq) K a = 1.0x10

14/3.8x10 10 = 2.6x10 5

HOCl (aq) + H 2O (l) H 3O+ (aq) + OCl (aq) K a =3.5x10

8 C6H5 NH 2 (aq) + H 3O

+ (aq) C 6H5 NH 3+ (aq) + H 2O (l) K = 1/2.6x10

5 = 3.8x10 4 _______________________________________________________________________

HOCl (aq) + C 6H5 NH 2 (aq) OCl (aq) + C 6H5 NH 3

+ (aq) K = 3.5x10 8 x 3.8x10 4 = 1.3x10 3

An important result is that K


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In general, only the strongest acid will significantly contribute to the concentration of hydronium ions.

(1) Strong Acid/Weak Acidex. Calculate the pH of a mixture containing 0.50 M HCl and 0.50 M HOCl.Since HCl is a strong acid and HOCl is a weak acid, the HOCl will not significantly contribute to the H + concentration, so the pHcan be calculated from the HCl concentration.

HCl H + + Cl 0.50 M 0.50 M

[H +] = 0.50 M pH = log[H +] = log(0.50) = 0.30

(if an ICE table were shown to determine [H 3O+] contribution from 0.50 M HOCl, the value obtained be 0.00013 M. This wouldnot change the value obtained for pH.)

(2) Two Weak Acidsex. Calculate the pH of a mixture containing 0.10 M HNO 2 (K a = 4.0x10

4) and 0.10 M HCN (K a = 6.2x10 10)

Since K a for HNO 2 >> than K a for HCN, only the HNO 2 will significantly contribute to [H 3O+]. An ICE table can then be used

to determine pH.

HNO 2 (aq) H2O (l) H3O+ (aq) + NO 2

(aq)

I 0.10 0 0

C x +x +x

E0.10 x 0.10assume xis small

x x

3 2

2

[ ][ ][ ]a

H O NO K

HNO

244.0 10

0.10 x

x

[H 3O+] = x = 6.3x10 3 M pH = log[H 3O

+] = log(6.3x10 3) = 2.20

Mixtures of BasesIn general, only the strongest base will significantly contribute to the concentration of hydroxide ions.

(1) Strong Base/Weak Baseex. Calculate the pOH of a mixture containing 0.20 M Sr(OH) 2 and 0.20 M NH 3.Since Sr(OH) 2 is a strong base and NH 3 is a weak base, the NH 3 will not significantly contribute to the OH

concentration, so the pOH can be calculated from the Sr(OH) 2 concentration.

Sr(OH) 2 Sr 2+ + 2OH

0.20 M 2(0.20 M)

= 0.40 M

[OH ] = 0.40 M pOH = log[OH ] = log(0.40) = 0.40

(if an ICE table were shown to determine [OH ] contribution from 0.20 M NH 3, the value obtained be 0.0019 M. This would notchange the value obtained for pOH.)

(2) Two Weak Basesex. Calculate the pOH of a mixture containing 0.50 M C 6H5 NH 2 (K b = 3.8x10

10) and 0.50 M NH 3 (K b = 1.8x10 5)

Since K b for NH 3 >> than K b for C 6H5 NH 2, only the NH 3 will significantly contribute to [OH ]. An ICE table can then be used

to determine pOH.

NH 3 (aq) H2O (l) OH (aq) + NH 4

+ (aq)

I 0.50 0 0

C x +x +x


x x

4

3

[ ][ ][ ]b

OH NH K

NH

251.8 100.50 x x

[OH ] = x = 3.0x10 3 M pOH = log[OH ] = log(3.0x10 3) = 2.52

*When in question about if a substance in a mixture will contribute significantly to [H +] or [OH ], it is always possible tocalculate the [H 3O

+] contribution from both acids and then add to determine the total [H 3O+] and use this value calculate pH OR

to calculate the [OH ] contribution from both bases and then add to determine the total [OH ] and use this value to calculate pOHand then pH.


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Solutions of Acids or Bases Containing a Common Ion of the Conjugate Base/AcidAn ICE table can be used to determine pH/pOH, where the initial concentration of the conjugate ion is not zero.

ex. A solution contains 1.0 M HF and 1.0 M NaF, Calculate the pH of the resulting solution.How does this pH compare to a 1.0 M solution of HF?

NaF Na + + F (note: NaF is a soluble salt)1.0 M 1.0 M the initial [F ]

HF (aq) H2O (l) H3O+ (aq) + F (aq)

I 1.0 0 1.0

C x +x +x


x1.0 + x 1.0assume xis small

3[ ][ ][ ]a

H O F K HF

4 (1.0)7.2 101.0

x x

[H 3O+] = x = 7.2x10 4 M

pH = log[H 3O+] = log(7.2x10 4) = 3.14

The pH of 1.0 M HF in water is 1.57 (calculated earlier this unit); the pH of 1.0 M HF in 1.0 M NaF is 3.14 Note : the pH of 1.0 M HF containing a common ion (F ) is higher than the pH of 1.0 M HF in water.Increasing [F ] shifts the equilibrium to the left, thereby decreasing [H 3O

+], making the solution LESS acidic and increasing pH.

ex. A solution contains 1.0 M NH 3 and 1.0 M NH 4Cl, Calculate the pH of the resulting solution.How does this pH compare to a 1.0 M solution of NH 3?

NH 4Cl NH 4+ + Cl (note: NH 4Cl is a soluble salt)

1.0 M 1.0 M the initial [NH 4+]

NH 3 (aq) H2O (l) OH (aq) + NH 4

+ (aq)

I 1.0 0 1.0

C x +x +x

E

1.0 x 1.0

assume xis small x

1.0 + x 1.0

assume xis small

4

3

[ ][ ][ ]b

OH NH K

NH

5 (1.0)1.8 101.0

x x

[OH ] = x = 1.8x10 5 M

pOH = log[OH ] = log(1.8x10 5) = 4.74

pH = 14.00pOH = 14.00 4.74 = 9.26

The pH of 1.0 M NH 3 in water is 11.62 (calculated earlier this unit); the pH of 1.0 M NH 3 in 1.0 M NH 4Cl is 9.26 Note : the pH of 1.0 M NH 3 containing a common ion (NH 4

+) is lower than the pH of 1.0 M NH 3 in water.Increasing [NH 4

+] shifts the equilibrium to the left, thereby decreasing [OH ], making the solution LESS basic and decreasing pH.


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Number of Acidic Protons/Hydrogen atommonoprotic acids : have only one acidic hydrogen atoms (ex. HCl, HNO 2, and HC 2H3O2)(Some compounds have formulas with more than one hydrogen atom; however, only one hydrogen atom can actually be donatedeasily in an acid equilibrium. This proton is sometimes referred to as the "labile" proton.)diprotic acids : have two acidic hydrogen atoms (ex. H 2SO 4 and H 2C2O4)polyprotic acids : have three or more acidic hydrogen atoms (ex. H 3PO 4)(diprotic and polyprotic acids have more than one labile proton.)

ex. Successive ionization of carbonic acid (H 2CO 3).H2CO 3 + H 2O (l) H 3O

+ (aq) + HCO 3 (aq) K a1 = 4.3x10

7 HCO 3

+ H 2O (l) H 3O+ (aq) + CO 3

2 (aq) K a2 = 5.6x10 11

_______________________________________________________ H2CO 3 + 2H 2O (l) 2H 3O

+ (aq) + CO 32 (aq) K = K a1 x K a2 = 4.3x10

7 x 5.6x10 11 = 2.4x10 17

For a diprotic acid generally, K a1>K a2. (The first proton is more easily lost than the second proton.)

AcidBase Properties of Salts The cation or anion of a salt may hydrolyze (react with water) and produce an acidic or basic solution.

CationsCations found in strong bases will not hydrolyze.

Cations that are the conjugate acids of weak bases will hydrolyze to produce acidic solutions.

AnionsAnions found in strong acids will not hydrolyze.Anions that are the conjugate bases of weak acids will hydrolyze to produce basic solutions.

ex. Determine if the pH of the following solutions will be acidic, basic, or neutral.(1) NaCl

NaOH is a strong base so Na + will not hydrolyzeHCl is a strong acid so Cl will not hydrolyzeSince neither the cation or anion hydrolyze, the solution will be neutral

(2) NaC 2H3O2 NaOH is a strong base so Na + will not hydrolyzeC2H3O2

is the conjugate base of the weak acid HC 2H3O2, so C 2H3O2 will hydrolyze:

C2H3O2 (aq) + H 2O (l) OH

(aq) + HC 2H3O2 (aq)The resulting hydrolysis reaction produces hydroxide ions so the solution is basic.

(3) NH 4ClHCl is a strong acid so Cl will not hydrolyze

NH 4+ is the conjugate acid of the weak base NH 3, so NH 4

+ will hydrolyze: NH 4

+ (aq) + H 2O (l) H 3O+ (aq) + NH 3 (aq)

The resulting hydrolysis reaction produces hydronium ions so the solution is acidic.

(4) NH 4 NO 2 NO 2

is the conjugate base of the weak acid HNO 2, so NO 2 will hydrolyze:

NO 2

(aq) + H 2O (l) OH

(aq) + HNO 2 (aq)

NH 4+ is the conjugate acid of the weak base NH 3, so NH 4

+ will hydrolyze: NH 4

+ (aq) + H 2O (l) H 3O+ (aq) + NH 3 (aq)

Since both ions hydrolyze, the K a and K b values for the resulting equilibria must be compared.

HNO 2 has K a = 4.0x10 4, therefore, NO 2

has K b = 2.5x10 11

NH 4+ has a K a = 5.6x10

10

Since K a > K b, the hydrolysis of NH 4+ will favour the products more than the hydrolysis of NO 2

, as a result there will be morehydronium ions than hydroxide ions and the solution will be acidic.


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Calculations for Saltsex. Calculate the pH and pOH of a 0.20 M solution of NaF.The concentration of each ion can be found from dissociation (since NaF is a soluble salt)

NaF Na + + F 0.20 M 0.20 M 0.20 M

NaOH is a strong base so Na + will not hydrolyze.F is the conjugate base of the weak acid HF , so F

will hydrolyze:F (aq) + H 2O (l) OH

(aq) + HF (aq) K b = K w/K a = 1.0x10 14/7.2x10 4 = 1.4x10 11

An ICE table can be used to determine the pOH of the solution.

F (aq) + H2O (l) OH (aq) + HF (aq)

I 0.20 0 0

C x +x +x

E

0.20 x0.20

assume xis small

x x

[ ][ ][ ]b

OH HF K

F

2111.4 10

0.20 x

x

[OH ] = x = 1.7x10 6 M pOH = log[OH ] = log(1.7x10 6) = 5.77 pH = 14.00pOH = 14.00 5.77 = 8.23

ex. Calculate the pH and pOH of a 0.015 M solution of NH 4Br.The concentration of each ion can be found from dissociation (since NH 4Br is a soluble salt) NH 4Br NH 4

+ + Br 0.015 M 0.015 M 0.0150 M

HBr is a strong acid so Br will not hydrolyze. NH 4

+ is the conjugate acid of the weak base NH 3, so NH 4+ will hydrolyze:

NH 4+ (aq) + H 2O (l) H 3O

+ (aq) + NH 3 (aq) K a = 5.6x10 10 (from K a chart)

An ICE table can be used to determine the pH of the solution.

NH 4+ (aq) + H2O (l) H3O

+ (aq) + NH 3 (aq)

I 0.015 0 0

C x +x +x

E

0.015 x 0.015

assume xis small

x x

3 3

4

[ ][ ][ ]a

H O NH K

NH

2105.6 10

0.015 x x

[H 3O+] = x = 2.9x10 6 M

pH = log[H 3O+] = log(2.9x10 6) = 5.54

pOH = 14.00pH = 14.00 5.54 = 8.46

ex. Calculate the pH and pOH of a 0.12 M solution of ethylammonium nitrite (C 2H5 NH 3 NO 2).

The concentration of each ion can be found from dissociation (C 2H5 NH 3 NO 2 is a soluble salt)C2H5 NH 3 NO 2 C 2H5 NH 3

+ + NO 2

0.12 M 0.12 M 0.12 M

NO 2 is the conjugate base of the weak acid HNO 2, so NO 2

will hydrolyze:

NO 2 (aq) + H 2O (l) OH (aq) + HNO 2 (aq) K b = K w/K a = 1.0x10 14/4.0x10 4 = 2.5x10 11

C2H5 NH 3+ is the conjugate acid of the weak base C 2H5 NH 2, so C 2H5 NH 3

+ will hydrolyze:C2H5 NH 3

+ (aq) + H 2O (l) H 3O+ (aq) + C 2H5 NH 2 (aq) K a = K w/K b = 1.0x10

14/4.4x10 4 = 2.3x10 11

Since both ions hydrolyze, and ICE table must be used to determine the [OH ] resulting from the hydrolysis of NO 2 and the

[H 3O+] from the hydrolysis of C 2H5 NH 3

+. The ion in excess can then be used to calculate the pH/pOH of the solution.


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NO 2 (aq) + H2O (l) OH (aq) + HNO 2 (aq)

I 0.12 0 0

C x +x +x


x x

2

2

[ ][ ][ ]b

OH HNO K

NO

2112.5 10

0.12 x

x

[OH ] = x = 8.2x10 6 M

C2H5 NH 3+

(aq) + H2O (l) H3O+

(aq) + C 2H5 NH 2 (aq)I 0.12 0 0

C x +x +x


x x

3 2 5 2

2 5 3

[ ][ ][ ]a

H O C H NH K C H NH

2112.3 10

0.12 x

x

[H 3O+] = x =1.6x10 6 M

Since K b for NO 2 >K a for C 2H5 NH 3

+, this results in [OH ]>[H 3O+]. The OH and H 3O

+ will partially neutralize each other,forming water (OH + H 3O

+ 2H 2O) with H 3O+ as the limiting reactant and OH as the excess reactant. The amount of excess

OH is found by subtracting the amount of the limiting reactant H 3O+.

[OH ] = 8.2x10 6 M 1.6x10 6 M = 6.6x10 6 M

pOH = log[OH ] = log(6.6x10 6) = 5.18 pH = 14.00pOH = 14.00 5.18 = 8.82

Neutralizations ReactionsTo give the net ionic equation for a neutralization reaction, the spectator ions must be determined. Only strong acids, strong

bases, and soluble salts will dissociate completely to form ions. Weak acids, weak bases, and water will remain undissociated.

There are three types of neutralization reactions that will be considered that will each result in a different form of net ionicequation.

(1) A Strong Acid with a Strong Baseex. A solution of hydrobromic acid is reacted with a solution of potassium hydroxide.HBr (aq) + KOH (aq) H2O (l) + KBr (aq)

H+ (aq) + Br (aq) + K + (aq) + OH (aq) H 2O (l) + K + (aq) + Br (aq) (the spectator ions are K + and Br )

H+ (aq) + OH (aq) H2O (l)

The reaction between all strong acids and bases will have the net ionic equation: H + (aq) + OH (aq) H 2O (l)

(2) A Weak Acid with a Strong Baseex. A solution of hydrofluoric acid is reacted with a solution of sodium hydroxideHF (aq) + NaOH (aq) H2O (l) + NaF (aq)

HF (aq) + Na + (aq) + OH (aq) H 2O (l) + Na+ (aq) + F (aq) (the spectator ion is Na +)

HF (aq) + OH (aq) H 2O (l) + F (aq)

The reaction between all weak acids and strong bases will have the general net ionic equation:HA (aq) + OH (aq) H 2O (l) + A

(aq)

(3) A Weak Base with a Strong Acidex. a solution of nitric acid reacts with a solution of ammonia. For this type of reaction, it is easiest to consider the strong acid asequivalent to hydronium, H 3O

+ (When HNO 3 is present in solution, it dissociates completely to H+ and NO 3

. Nitrate, NO 3 , is a

spectator ion, but the hydrogen ion, H +, will combine with H 2O and result and form hydronium, H 3O+. This is why the hydrogen

ion and hydronium ion are considered to be equivalent). The weak base in the reaction (i.e. NH 3) will accept a proton from H 3O+

to produce water. NH 3 (aq) + H 3O

+ (aq) H2O (l) + NH 4+ (aq)

The reaction between all weak bases and strong acids will have the general net ionic equation:B (aq) + H 3O

+ (aq) H 2O (l) + BH+ (aq)


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BuffersBuffer : a solution that resists changes in pH.

A buffer contains a weak acid (HA) and (a salt of) its conjugate base (A ).HA (aq) + H 2O (l) H 3O

+ (aq) + A (aq)

OR a weak base (B) and (a salt of) its conjugate acid (BH +)B (aq) + H 2O (l) OH

(aq) + BH + (aq)

Examples of Buffer Solutions:

(1) Example of an acidic buffer: HC 2H3O2 and NaC 2H3O2 (provides C 2H3O2 )

equilibrium: HC 2H3O2 (aq) + H 2O (l) H 3O+ (aq) + C 2H3O2

(aq)

(2) Example of a basic buffer: NH 3 and NH 4Cl (provides NH 4+)

equilibrium: NH 3 (aq) + H 2O (l) OH (aq) + NH 4

+ (aq)

Examples of Solutions that are NOT Buffers(1) HCl and NaClSince HCl is a strong acid this solution would not act as a buffer. Only weak acids and their conjugate bases (or weak bases andtheir conjugate acids) make good buffers. Strong acids (and bases) do not make good buffers. It takes much more base to change

the pH of a weak acid solution because there is a large amount of undissociated weak acid.

(2) H 2SO 3 and Na 2SO 3 Since SO 3

2 is not the conjugate base of H 2SO 3 this solution would not act as a buffer. In order to prepare a buffer using H 2SO 3,the solution would need to contain HSO 3

. A buffer solution could be prepared by combining H 2SO 3 and NaHSO 3.

Buffer CalculationsBuffer calculations are equivalent to those for solutions with a common ion and can be solved using an ICE Table.

ex. A solution contains 0.50 M HC 2H3O2 and 0.50 M NaC 2H3O2. Calculate the pH of the solution.

HC 2H3O2 (aq) + H 2O (l) H 3O+ (aq) + C 2H3O2

(aq)

I 0.50 0 0.50

C x +x +x

E

0.50 x0.50

assume xis small

x

0.50 + x0.50

assume xis small

3 2 3 2

2 3 2

[ ][ ][ ]a

H O C H O K HC H O

5 (0.50)1.8 100.50

x x

[H 3O+] = x = 1.8x10 5 M

pH = log[H 3O+]

= log(1.8x10 5)= 4.74

The HendersonHasselbalch Equation Determining the pH of any solution containing a weak acid and its conjugate base can be simplified using the Hendersen-Hasselbalch equation:

[ ]log

[ ]a A

pH pK HA

where: pK a = log K a

Determining the pOH of any solution containing a weak base and its conjugate acid can be simplified using the followingequation:

[ ]log

[ ]b BH

pOH pK B

where: pK b = log K b


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ex. A buffer solution contains 0.50 M HC 2H3O2 and 0.50 M NaC 2H3O2. Calculate the pH of the solution.

2 3 2

2 3 2

[ ]log

[ ]aC H O

pH pK HC H O

5 0.50log(1.8 10 ) log0.50

pH x

4.74 pH (as previously calculated using an ICE table)

note: When [HA] = [A

] (in this case, [HC 2H3O2] = [C 2H3O2

]), then pH = pK a.


2 3 2

2 3 2

[ ]log

[ ]aC H O

pH pK HC H O

5 0.10log(1.8 10 ) log0.50

pH x

4.05 pH

note: When [HA] > [A ] (in this case, [HC 2H3O2] > [C 2H3O2 ]), then pH < pK a.


2 3 2

2 3 2

[ ]log

[ ]aC H O

pH pK HC H O

5 0.50log(1.8 10 ) log0.10

pH x

5.44 pH

note: When [HA] < [A

] (in this case, [HC 2H3O2] < [C 2H3O2

]), then pH > pK a.

ex. A solution contains 0.25 M NH 3 and 0.30 M NH 4Cl . Calculate the pH of the solution.

4

3

[ ]log

[ ]b NH

pOH pK NH

5 0.30log(1.8 10 ) log0.25

pOH x

4.82 pOH

pH = 14.00 pOH = 14.00 4.05 = 9.18


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The mechanism of a buffer upon addition of acid or base can be explained using the HendersonHasselbalch Equation

[ ]log

[ ]a A

pH pK HA

Consider an the following buffer: HA (aq) + H 2O (l) H 3O+ (aq) + A (aq)

(1) If a base, OH , is added to the buffer, the base will react with the acidic component of the buffer:HA (aq) + OH (aq) H2O (l) + A

(aq)

As a result, [A ] will increase and [HA] will decrease (and the ratio of [A ]/[HA] will increase). Therefore, the addition of baseresults in an overall slight increase in pH.

(2) If an acid, H 3O+, is added to the buffer, the acid will react with the basic component of the buffer:

A (aq) + H 3O+ (aq) H 2O (l) + HA (aq)

As a result, [HA] will increase and [A ] will decrease (and the ratio of [A ]/[HA] will decrease). Therefore, the addition of acidresults in an overall slight decrease in pH.

Buffer CapacityBuffer Capacity: the amount of an acid or base that can be added to a specific volume of a buffer solution before its pH changes

significantly (usually defined as an increase or decrease of one pH unit).A buffer is most effective at resisting change in pH near its pK a value. If a buffer has pH = pK a the ratio [A ]/[HA] = 1. If too

much acid or base is added to the buffer, the ratio of [A ]/[HA] will be altered such that the pH will change significantly. When base is added, the ratio can be increased such that [A ]/[HA] changes to 10 and the pH will increase by one; when acid is added,the ratio can be decreased such that [A ]/[HA] changes to 0.10 and the pH will decrease by one.Buffer capacity is also determined by the amount of acid and of conjugate base. The more moles of each that are present, thehigher the buffer capacity. For buffers of equal volume, the greater the concentrations of both HA and A , the greater the

buffering capacity since more acid or base would need to be added in order to change the ratio of [A ]/[HA] to a value thatwould significantly change the pH. A solution with high [HA] and [A ] will be able to more effectively resist changes in pH.For buffers of equal concentrations of HA and A , the higher the volume, the greater the buffer capacity. The more volume, themore moles of both HA and A that would be present and the more acid or base would need to be added in order to change theratio of [A ]/[HA] to a value that would significantly change the pH.

Preparing BuffersIn order to calculate the pH of a buffer, the pK a (or pK b) must be known along with the concentration of weak acid and conjugate

base (or the concentration of weak base and conjugate acid).

ex. Calculate the pH of a buffer made by adding 0.75 g of sodium fluoride to 200 mL of 0.10 M hydrofluoric acid. Assume thevolume of the solution remains constant.

ex. Determine the mass of methylammonium chloride (CH 3 NH 3Cl) that should be added to 300 mL of 0.20 M methylamine(CH 3 NH 2) in order to make a buffer with a pH of 10.60. Assume the volume of the solution remains constant.
http://c/Users/Sarah/Desktop/acid.htmlhttp://c/Users/Sarah/Desktop/base.htmlhttp://c/Users/Sarah/Desktop/buffsol.htmlhttp://c/Users/Sarah/Desktop/ph.htmlhttp://c/Users/Sarah/Desktop/ph.htmlhttp://c/Users/Sarah/Desktop/buffsol.htmlhttp://c/Users/Sarah/Desktop/base.htmlhttp://c/Users/Sarah/Desktop/acid.html


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Biological Applications: The Blood BufferThe body maintains the pH of blood between 7.35 and 7.45. If the pH level changes just a few tenths of a pH unit, serious healthconsequences can result. A decrease in blood pH is called acidosis , an increase is called alkalosis .The most important buffer in the blood consists of a buffer between carbonic acid (H 2CO 3) and hydrogen carbonate (HCO 3

, alsocalled bicarbonate)H2CO 3 (aq) + H 2O (l) H 3O

+ (aq) + HCO 3 (aq)

In addition, there is a phosphate buffer (H 2PO 4 /HPO 4

2 ) as well as the buffering ability of some proteins in plasma (since theyhave a carboxylic acid/amine i.e. a weak acid/base component) that are available to maintain blood pH.

TitrationsTitration : An analytical procedure in which the concentration of a known solution is used to determine the concentration of anunknown solution.

Acid Base TitrationsAcid base titrations involve neutralization reactions. For an experiment, the objective is to determine the volume of a solutionwith known concentration that is required to titrate (i.e. exactly react with or neutralize) a given volume of solution with anunknown concentration. The solution with known concentration is called the titrant . The solution with the unknownconcentration is called the titrate or the analyte . The higher the concentration of the titrate/analyte, the more of the titrant thatwill be required in order to complete the neutraliz ation reaction. Usually several trials are performed and the average volumeis used for calculations. Chemical indicators (i.e. phenolphthalein or bromothymol blue) can be used to monitor when thereaction is complete and has reached its equivalence point.

Titration CalculationsDetermine the balanced chemical equation for the neutralization reaction. Titration calculations are solution stoichiometry

problems. Use the concentration and volume given to determine the moles of titrant used in the experiment. Determine themoles of titrate/analyte using the mole ratio in the reaction. Determine the concentration of the titrate/analyte from the volume.

ex. What is the concentration if 12.00 mL of HCl requires 15.00 mL of 0.20 M NH 3 to react completely?

H3O+ (aq) + NH 3 (aq) H2O (l) + NH 4

+ (aq) moles (mol) 0.0030 0.0030

mass (g)

volume (L) 0.01200 0.01500Molarity (M) 0.25 0.20molar mass

(g/mol)

= = (0.20 M) (0.01500 L) = 0.0030 mol NH 3

33 3

3

10.0030 0.0030 0.0030

1mol H O

mol NH x mol H O mol HCl mol NH

= 0.0030

[ ] 0.250.01200mol

HCl M L


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ex. What is the concentration if 5.00 mL of HF requires 25.00 mL of 0.10 M Ba(OH) 2 to react completely?

Ba(OH) 2 Ba2+ + 2OH

therefore, [OH ] = 2(0.10 M) = 0.20 M

HF (aq) + OH (aq) H2O (l) + F (aq)

moles (mol) 0.0050 0.0050

mass (g)

volume (L) 0.00500 0.02500

Molarity (M) 1.0 0.20molar mass

(g/mol)

= = (0.20 M) (0.02500 L) = 0.0050 mol OH

10.0050 0.0050

1mol HF

mol OH x mol HF mol OH

= 0.0050

[ ] 1.00.00500mol

HF M L


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Titration Curve CalculationsIn a titration experiment, the pH of the solution at incremental points in the titration experiment can be determined and graphedto create a titration curve.

Strong Acid/ Strong Base Titrationex. A 0.100 M solution of NaOH is used to titrate 100.0 mL of a 0.100 M HCl solution.

Determine the pH at the following intervals in the experiment.(1) No NaOH has been added pH = 1.000The pH of the solution can be calculated from the [HCl]

HCl H + + Cl

therefore, [H+] = 0.100 M pH = log[H +]= log(0.100)= 1.000

(2) 25.00 mL of NaOH has been added pH = 1.22The acid and base will partially neutralize each other and the

pH of the resulting solution can be determined from theconcentration of the excess reactant.


initialmol

= (0.100M)

(0.1000 L)= 0.0100 molH+ is the EXCESSreactant

= (0.100 M)(0.02500 L)= 0.00250 molOH is theLIMITINGreactant

It is notnecessary

to calculatethe molesof water

producedchangein mol 0.00250 mol 0.00250 mol

finalmol

= 0.0075 mol

= 0 (since OH isthe limitingreactant, it iscompletelyconsumed)

0.0075[ ] 0.060

0.1250mol

H M L

(note: the total volume of the solution is0.1000 L + 0.02500 L = 0.1250 L)

pH = log[H +] = log(0. 060) = 1.22

(3) 50.00 mL of NaOH has been added pH = 1.48




initialmol

= (0.100M)(0.1000 L)= 0.0100 mol

= (0.100M)(0.1000 L)= 0.0100 mol

changein mol 0.0100 mol 0.0100 mol

finalmol = 0 = 0

Since only H 2O is present, the solution is neutral with pH =7.00. This point is call the equivalence point because molesH+ = moles OH (moles acid = moles base)



initialmol

= (0.100M)(0.1000 L)= 0.0100 molH+ is theLIMITINGreactant

= (0.100 M)(0.12500 L)= 0.01250 molOH is theEXCESSreactant

changein mol 0.0100 mol 0.0100 mol

finalmol

= 0 (since H + is thelimiting reactant, itis completelyconsumed)

= 0.0025 mol

0.0025[ ] 0.011

0.2250mol

OH M L

(note: the total volume of the solution is0.1000 L + 0.12500 L = 0.2250 L)

pO H = log[ OH ] = log(0. 011) = 1.96 pH = 14.00 pH = 14.00 1.96 = 12.04



(9) 200.0 mL of NaOH has been added. pH = 12.52

Note that there are four regions to be considered in titration curve calculations where a strong base is added to a strong acid:(1) Before any base is added- pH is calculated directly from the concentration of acid.(2) Before the equivalence point- pH is calculated from the excess concentration of acid.(3) At the equivalence point- since moles H + = moles OH , pH = 7(4) After the equivalence point- pOH (and then pH) is calculated from the excess concentration of base.


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The pH values for each volume of NaOH added can be graphed to create a titration curve.

If a strong acid is added to a strong base, the four regions to be considered in titration curve calculations are as follows:(1) Before any acid is added- pOH (and then pH) is calculated directly from the concentration of base.(2) Before the equivalence point- pOH (and then pH) is calculated from the excess concentration of base.(3) At the equivalence point- since moles acid (H +) = moles base (OH ), pH = 7(4) After the equivalence point- pH is calculated from the excess concentration of acid.

Conductivity in TitrationsA titration can also be performed by measuring the conductivity of the solution (the solution being titrated will conductelectricity due to the presence of ions). This is called a conductometric titration.

Strong Acid/Strong Base (i.e. HCl and NaOH)As NaOH is added, the concentration of H + ions decreases (H + + OH H 2O)and the concentration of Na + ions is increased. There is a constant amount ofCl ion in the solution since it is a spectator ion. The total charge of thesolution is zero. Although the positive charge of the H + ions is replaced by

Na + ions, the conductivity of an H + ion is greater than that of an Na + ion, so theoverall conductivity of the solution decreases. At the equivalence point, all ofthe H + and OH have been consumed, so the conductivity of the solution is at aminimum (since the solution contains only Na + and Cl ). After the equivalence

point, if additional NaOH is added, the conductivity increases due to therelatively higher conductivity of the OH ions in solution.

0123456789

1011121314

0 25 50 75 100 125 150 175 200

p H

Volume NaOH added (mL)

Strong Acid Strong Base Titration


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Strong Base/Weak Acidex. A 0.050 M solution of NaOH is used to titrate 100.0 mL of a 0.050 M benzoic acid (HC 7H5O2, K a = 6.3x10

5) solution.

Determine the pH at the following intervals in the experiment.(1) No NaOH has been added pH = 2.75

Before any base has been added, the pH can be calculated using an ICE table.HC 7H5O2 (aq) + H2O (l) H3O

+ (aq) + C 7H5O2 (aq)

I 0.050 0 0

C x +x +x

E 0.050 x x x


3 7 5 2

7 5 2

[ ][ ][ ]a

H O C H O K

HC H O

256

0.050.3 10

x x

x = 0.0018 M = [H 3O+]

pH = log[H 3O+]

= log(0.0018)= 2.75


(3) 50.00 mL of NaOH has been added pH = 4.20The added base will partially neutralize the acid. Determine moles/molarity of excess acid and conjugate base produced in thereaction. The pH of the solution can be found using an ICE table OR using the HendersonHasselbalch Equation .

HC 7H5O2 (aq) + OH (aq) H2O (l) + C7H5O2 (aq)

initialmol

= (0.050M)(0.1000 L)= 0.0050 molHC 7H5O2 is theEXCESS reactant

= (0.050M)(0.0500 L)= 0.0025 molOH is the LIMITINGreactant

It is notnecessary

to calculate themoles of water

produced

0(initially, there is no

conjugate base present)

changein mol 0.0025 mol 0.0025 mol + 0.0025 mol

final mol = 0.0025 mol= 0 (since OH is thelimiting reactant, it iscompletely consumed)

= 0.0025 mol

7 5 20.0025[ ] 0.017

0.1500mol HC H O M L

7 5 2 0.0025[ ] 0.0170.1500mol C H O M L

(note: the total volume of the solution is 0.1000 L + 0.05000 L = 0.1500 L)

HC 7H5O2 (aq) + H2O (l) H3O+ (aq) + C 7H5O2

(aq)

I 0.017 0 0.017

C x +x +x

E

0.017 x 0.017

assume xis small

x

0.017 + x 0.017

assume xis small

3 7 5 2

7 5 2

[ ][ ][ ]a

H O C H O K

HC H O

5 (0.017)6.3 10.017

0 x

x

x = 6.3x10 5 M = [H 3O+]

pH = log[H3O+]

= log(6.3x10 5)= 4.20

OR

HendersonHasselbalch Equation

7 5 2

7 5 2

[ ]log

[ ]aC H O

pH pK HC H O

5

0.017log(6.3 10 ) log

0.017 pH x

pH = 4.20 (as with the ICE table)



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initialmol

= (0.050M)(0.1000 L)= 0.0050 mol

= (0.050M)(0.1000 L)= 0.0050 mol

It is notnecessary


produced




final mol = 0 = 0 = 0.0050 mol

This is the equivalence point in the titration because moles acid = moles base.The only substance present at the equivalence point is the conjugate base.

7 5 2

0.0050[ ] 0.025

0.2000mol

C H O M L


For a Weak Acid/Strong Base Titration, the solution is NOT neutral at the equivalence point. The conjugate base will form anequilibrium with water and the pOH (and then the pH) can be calculated using an ICE table.

C7H5O2 (aq) + H2O (l) OH

(aq) + HC 7H5O2 (aq)

I 0.025 0 0

C x +x +x

E

0.025 x0.025

assume xis small

x x

7 5 2

7 5 2

[ ][ ][ ]b

OH HC H O K

C H O

(the K n expression and constant is usedsince there is a base equilibrium)

1410

5

1.0 101.6 10

6.3 10w

ba

K x K x

K x

2101.6 10

0.025 x

x

x = 2.0x10 6 M = [OH ]

pOH = log[OH ] = log(2.0x10 6) = 5.70 pH = 14.00 pOH = 14.00 5.70 = 8.30




initialmol

= (0.050M)(0.1000 L)= 0.0050 molHC 7H5O is theLIMITING reactant

= (0.050M)(0.1500 L)= 0.0075 molOH is the EXCESSreactant

It is notnecessary


produced




final mol

= 0(since HC 7H5O2 is thelimiting reactant, it iscompletely consumed)

= 0.0025 = 0.0050 mol

The solution contains a mixture of excess strong base (OH ) and conjugate base (C 7H5O2 ). The pOH (and then the pH) can becalculated from the molarity of strong base because the weak conjugate base will not significantly contribute to [OH ].

0.0025[ ] 0.010

0.2500mol

OH M L

pOH = log[OH ] = log(0.010) = 2.00 pH = 14.00 pOH = 14.00 2.00 = 12.00



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(8) 175.0 mL of NaOH has been added pH =12.13

(9) 200.0 mL of NaOH has been added. pH = 12.22

The pH values for each volume of NaOH added can be graphed to create a titration curve.

If a strong base is added to a weak acid, the four regions to be considered in titration curve calculations are as follows:(1) Before any base is added- pH is calculated from the concentration and K a of the acid using an ICE table.(2) Before the equivalence point- pH is calculated from the remaining concentration of acid and produced concentration ofconjugate base using an ICE table OR using the Henderson-Hasselbalch equation.(3) At the equivalence point- pOH (and then pH) is calculated from the concentration of conjugate base using an ICE table andK b.(4) After the equivalence point- pOH (and then pH) is calculated from the excess concentration of base.

Strong Acid/Weak Baseex. A 0.100 M solution of HCl is used to titrate 100.0 mL of a 0.100 M NH 3 (K b = 1.8x10

5) solution.

Determine the pH at the following intervals in the experiment.(1) No HCl has been added pH = 11.11Before any acid has been added, the pOH (and then the pH) can be calculated using an ICE table.

NH 3 (aq) + H 2O (l) OH (aq) + NH 4

+ (aq)

I 0.100 0 0

C x +x +xE 0.100 x x x


4

3

[ ][ ][ ]b

OH NH K

NH

251.8 10

0.100

x x

x = 0.0013 M = [OH ]

pOH = log[OH ] = log(0.0013) = 2.89

pH = 14.00 pOH = 14.00 2.89 = 11.11

0123456789

10111213

14

0 25 50 75 100 125 150 175 200

p H

Volume NaOH added (mL)

Weak Acid Strong Base Titration


25/30

(2) 25.00 mL of HCl has been added pH = 9.73The added acid will partially neutralize the base. Determine moles/molarity of excess base and conjugate acid produced in thereaction. The pOH (and then pH) of the solution can be found using an ICE table OR using the HendersonHasselbalchEquation.

NH 3 (aq) + H 3O+ (aq) H2O (l) + NH 4

+ (aq)

initialmol

= (0.100M)(0.1000 L)= 0.0100 mol

NH 3 is theEXCESS reactant

= (0.100M)(0.02500 L)= 0.00250 molH3O

+ is theLIMITINGreactant

It is notnecessary

to calculatethe molesof water

produced

0(initially, there isno conjugate acid

present)


finalmol

= 0.0075 mol

= 0 (since H 3O+ is

the limitingreactant, it iscompletelyconsumed)

= 0.0025 mol

30.0075

[ ] 0.0600.1250

mol NH M

L 4

0.0025[ ] 0.020

0.1250mol

NH M L


ICE TABLE

NH 3 (aq) + H 2O (l) OH (aq) + NH 4

+ (aq)

I 0.060 0 0.020

C x +x +x

E

0.060 x 0.060

assume x issmall

x 0.020 + x 0.020

4

3

[ ][ ][ ]b

OH NH K

NH

5 (0.020)1.8 100.060

x x

x = 5.4x10 5 M = [OH ]

pOH = log[OH ] = log(5.4x10 5) = 4.27

pH = 14.00 pOH = 14.00 4.27 = 9.73

OR

HendersonHasselbalch Equation :

4

3

[ ]log

[ ]b NH

pOH pK NH

5

0.020log(1.8 10 ) log

0.060 pOH x

pOH = 4.27 (as with the ICE table)

(3) 50.00 mL of HCl has been added pH = 9.26

(4) 75.00 mL of HCl has been added pH = 8.77


26/30


27/30

The pH values for each volume of HCl added can be graphed to create a titration curve.

If a strong acid is added to a weak base, the four regions to be considered in titration curve calculations are as follows:(1) Before any acid is added- pOH (and then pH) is calculated from the concentration and K b of the base using an ICE table.(2) Before the equivalence point- pOH (and then pH) is calculated from the remaining concentration of base and producedconcentration of conjugate acid using an ICE table OR using the Henderson-Hasselbalch equation.(3) At the equivalence point- pH is calculated from the concentration of conjugate acid using an ICE table and K a.(4) After the equivalence point- pH is calculated from the excess concentration of acid.

Halfway to the Equivalence Point in a Titration Experiment Weak Acid/Strong BaseIn a titration experiment between a weak acid and a strong base, when the amount of base added is half the volume required toreach the equivalence point, the contents of the solution are the cation of the strong base (i.e. Na +), unreacted weak acid (HA),conjugate base (A ) that has been produced, and a small amount H 3O

+ from the acid equilibrium with water. At this point in thetitration, the total positive charge is equal to the total negative charge so the overall charge of the solution is (approximately)zero. The added OH has consumed half of the HA in the original solution by reacting to produce H 2O and A . The number ofmoles of the acid (HA) remaining will be equal to the number of moles of the conjugate base (A ) produced. In a given volume,the concentration of acid will be equal to the concentration of the conjugate base ([HA] = [A ]). Consider both the K a expression and the Henderson-Hasselbalch Equation at this point.

From the K a expression:3[ ][ ][ ]a

H O A K

HA

(since [HA] = [A ])

3[ ]a K H O

From the Henderson-Hasselbalch Equation: [ ]

log[ ]a A

pH pK HA

pH = pK a + log (1)(since [HA] = [A ], log 1 = 0)

pH = pK a If the pH halfway to the equivalence point is determined, the [H 3O

+] can be calculated, and therefore K a can be found, orequivalently, the pH of a solution halfway to the equivalence point will be equal to pK a.

Consider STEP 3 of the previous example (benzoic acid/sodium hydroxide). The volume of NaOH required to reach theequivalence point was 100.0 mL, therefore, half the volume required to reach the equivalence point was 50.00 mL. At this pointin the titration, the number of moles of HC 7H5O2 (acid remaining) was 0.0025 mol and the number of moles of C 7H5O2

(conjugate base produced) was also 0.0025 mol. The [HC 7H5O2] = [C 7H5O2

] = 0.017 M. Since both the acid and the conjugate base have the same concentration, pH = pKa (i.e. pH = log(6.3x10 5) = 4.20).

0123456789

1011121314

0 25 50 75 100 125 150 175 200

p H

Volume HCl added (mL)

Strong Acid Weak Base Titration


28/30

Summary of Titration CurvesTitration Initial pH

nearing theequivalence point

Change in pH aroundthe equivalence point

At theequivalence point Final pH Titration Curve

StrongAcid addedtoStrongBase

veryhigh(onlystrong

base is

present)

steadydecrease

extreme pHchange

pH = 7(water is

present)

very low(onlystrongacid is

present)

StrongBase addedtoStrongAcid

verylow(onlystrongacid is

present)

steady increase extreme pHchange

pH = 7(water is

present)

veryhigh(onlystrong

base is present)

StrongAcid addedtoWeak Base

mediumhigh(weak

base is present)

more gradualdecrease(buffer regionwith weak baseand conjugateacid )

more moderate pH change

pH7(conjugate

base is present)

veryhigh(primari

ly strong base is present)


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Strength of an acid and shape of the titration curveThe overall shape of a titration curve is determined by the strength (i.e. K a value) of the acid. The stronger an acid, the lower itsinitial pH; the weaker an acid the higher its initial pH. The stronger an acid, the lower the pH halfway to the equivalence point(since the pKa is lower); the weaker an acid, the higher the pH halfway to the equivalence point (since the pKa is higher).Stronger acids will have a more gradual initial increase in pH and then a more extreme increase in pH near the equivalence point;weaker acids will have a more noticeable initial increase in pH and then a less drastic increase in pH near the equivalence point.At the equivalence point, for stronger acids, the pH is lower (since the conjugate base is relatively weak); for weaker acids, the

pH is higher (since the conjugate base is relatively strong). After the equivalence point, the pH for both stronger acids an dweaker acids will be very similar since there is excess strong base in both cases.

Titration of 50.0 mL a 0.10 M solution of weak acid with 0.10 M NaOH

K a pK a

Initial pH of 0.10 Msolution

pH at volumehalfway to

equivalence point(25.0 mL NaOH

added)note: pK a = pH

pH at equivalencepoint

(50.0 mLNaOHadded)

10 2 1.6 2 7.3510 4 4 2.5 4 8.3510 6 3.5 6 9.3510 8 4.5 8 10.3510 10 10 5.5 10 11.35

Acid/Base IndicatorsThe equivalence point in an acid base titration is monitored using an acid base indicator. An indicator is a weak acid thatchanges colour depending on whether it is in its acidic form or its basic form.

ex. HInd (aq) + H 2O (l) H 3O+ (aq) + Ind (aq)

acidic form basic form

The point in a titration at which the indicator changes colour is called the endpoint. Every indicator changes colour at a specific pH value called its pK a value. When pHpK a for the indicator, the indicator will be in its basic form (Ind

) and will display thecolour of the basic form. An indicator is selected for a given titration because its pK a value is close to the pH at the equivalence

point.

ex. Bromothymol blue is yellow in its acidic form and blue in its basic form. Bromothymol blue has a pK a value of 7.0 andwould be useful for a titration with an equivalence point around 7 (i.e. a strong acid and a strong base)ex. Phenolphthalein is colourless in its acidic form and pink in its basic form. Phenolphthalein has a pK a value of 9.3 would beuseful for a titration with an equivalence point around 9 (i.e. a weak acid and a strong base)ex. Methyl red is red in its acidic form and yellow in its basic form. Methyl red has a pK a value of 5.1 and would be useful for atitration with an equivalence point around 5 (i.e. a strong acid and a weak base)

ex. Acidic and Basic forms for Phenolphthalein


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Indicators and the pH range of colour change. The middle of the range is the pKa for the indicator.

Equivalence Point vs. EndpointAn important distinction must be made between the equivalence point of a titration and the endpoint of a titration.The equivalent point, also called the stoichiometric point, is where the moles of acid are equal to the moles of base. At this pointof the experiment, the reaction is complete. The analyte is said to be titrated at this point. The endpoint of a titration is where the acid/base indicator is observed to change colour. The careful selection of an appropriateindicator is to ensure that the pK a of the indicator will be very close to the pH at the equivalence point (i.e. that the volumerequired to reach the equivalence point is a close as possible to the volume required to reach the endpoint).

Diprotic TitrationIf a strong base is titrated with a weak diprotic acid, there will be two equivalence points as shown in the following titrationcurve.

Consider the weak acid H 2A

The first equivalence point involves the reaction ofthe first proton in the acid with the added base.

H2A + OH H 2O + HA

The second equivalence point involves the reaction ofthe second proton in the acid with the added base.

HA + OH H 2O + A2

The total volume of base required to reach the secondequivalence point is twice the volume required toreach the first equivalence point.

For a triprotic (H 3A) acid, there would be three different equivalence points, one for each proton.

acids and bases note sap

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