scilearn.sydney.edu.au · chem1002 2014-n-2 november 2014 • calculate the ph of a 0.020 m...

CHEM1002 2014-N-2 November 2014

• Calculate the pH of a 0.020 M solution of lactic acid, HC3H5O3, at 25 °C. The pKa of lactic acid is 3.86.

5

As lactic acid is a weak acid, [H3O+] must be calculated using a reaction table:

HC3H5O3 H2O H3O+ C3H5O3-

initial 0.020 large 0 0

change -x negligible +x +x

final 0.020 –x large x x

The equilibrium constant Ka is given by:

Ka = 𝐇𝟑𝐎! [𝐂𝟑𝐇𝟓𝐎𝟑

!][𝐇𝐂𝟑𝐇𝟓𝐎𝟑]

= 𝒙𝟐

𝟎.𝟎𝟐𝟎!𝒙

As pKa = -log10Ka, Ka = 10−3.86 and is very small, 0.010 – x ~ 0.010 and hence:

x2 = 0.020 × 10–3.86 or x = 1.7 × 10-3 M = [H3O+]

Hence, the pH is given by:

pH = −log10[H3O+] = −log10(1.7 × 10-3) = 2.78

pH = 2.78

A 1.0 L solution of 0.020 M lactic acid is added to 1.0 L of 0.020 M sodium hydroxide solution. Write the ionic equation for the reaction that occurs.

HC3H5O3(aq) + OH-(aq) à C3H5O3

-(aq) + H2O(l)

Is the resulting solution acidic, basic or neutral? Give a reason for your answer.

All of the lactic acid will have reacted as the number of moles of sodium hydroxide added equals the number of moles of lactic acid originally present. The solution contains lactate ions, C3H5O3

-(aq), which is the conjugate base. The solution contains a weak base so is basic.


Calculate the pH of a 0.010 M solution of aspirin at 25 °C. The pKa of aspirin is 3.5 at this temperature.

Marks 7

As aspirin is a weak acid, [H3O+] must be calculated using a reaction table:

C9H8O4 H2O H3O+ C9H7O4-





Ka = 𝐇𝟑𝐎! [𝐂𝟗𝐇𝟕𝐎𝟒

!][𝐂𝟗𝐇𝟕𝐎𝟒]

= 𝒙𝟐

𝟎.𝟎𝟏𝟎!𝒙


x2 = 0.010 × 10–3.5 or x = 1.8 × 10-3 M = [H3O+]


pH = −log10[H3O+] = −log10(1.8 × 10-3) = 2.8

pH = 2.8

Aspirin, C9H8O4 is not very soluble. “Soluble aspirin” can be made by reacting aspirin with sodium hydroxide. Write the chemical equation for this reaction.

C9H8O4(s) + OH–(aq) → C9H7O4–(aq) + H2O(l)

Is a solution of “soluble aspirin” acidic or basic? Briefly explain your answer.

Basic. The C9H7O4–(aq) ion reacts with water (i.e. undergoes hydrolysis) to

generate a small amount of OH– ions. The C9H7O4–(aq) ion is a weak base, so the

following equilibrium reaction lies very much in favour of the reactants.

C9H7O4–(aq) + H2O(l) C9H8O4(aq) + OH–(aq)


• Solution A consists of a 0.050 M aqueous solution of benzoic acid, C6H5COOH, at 25 °C. Calculate the pH of Solution A. The pKa of benzoic acid is 4.20.

Marks 6

As benzoic acid is a weak acid, [H3O+] must be calculated using a reaction table:

C6H5COOH H+ C6H5COO-

initial 0.050 0 0

change -x +x +x

final 0.050 –x x x


Ka = 𝐇! [𝐂𝟔𝐇𝟓𝐂𝐎𝐎!][𝐂𝟔𝐇𝟓𝐂𝐎𝐎𝐇]

= 𝒙𝟐

𝟎.𝟎𝟓𝟎!𝒙


x2 = 0.050 × 10–4.2 or x = 1.78 × 10-3 M = [H+]


pH = −log10[H+] = −log10(1.78 × 10-3) = 2.75

pH = 2.75

What are the major species present in solution A?

Ka is very small and the equilibrium lies almost completely to the left. The major species present are water and the undissociated acid: H2O and C6H5COOH

Solution B consists of a 0.050 M aqueous solution of ammonia, NH3, at 25 °C. Calculate the pH of Solution B. The pKa of NH4

+ is 9.24.

NH3 is a weak base so [OH-] must be calculated by considering the equilibrium:

NH3 H2O NH4+ OH-


change -y negligible +y +y

final 0.050 – y large y y

ANSWER CONTINUES ON THE NEXT PAGE


The equilibrium constant Kb is given by:

Kb = 𝐍𝐇𝟒! [𝐎𝐇!]

[𝐍𝐇𝟑] =

𝒚𝟐

(𝟎.𝟎𝟓𝟎!𝒚)

For an acid and its conjugate base:

pKa + pKb = 14.00 pKb = 14.00 – 9.24 = 4.76

As pKb = 4.76, Kb = 10-4.76. Kb is very small so 0.050 – y ~ 0.050 and hence: y2 = 0.050 × 10–4.76 or y = 9.32 × 10-4 M = [OH-]

Hence, the pOH is given by: pOH = -log10[OH-] = log10[9.32 × 10-4] = 3.03

Finally, pH + pOH = 14.00 so pH = 14.00 – 3.03 = 10.97

pH = 10.97

What are the major species present in solution B?

Kb is very small and the equilibrium lies almost completely to the left. The major species present are water and the unprotonated weak base: H2O and NH3


Write the equation for the reaction that occurs when benzoic acid reacts with ammonia?

Marks 5

C6H5COOH(aq) + NH3(aq) → C6H5COO–(aq) + NH4+(aq)

Write the expression for the equilibrium constant for the reaction of benzoic acid with ammonia?

K = 𝐂𝟔𝐇𝟓𝐂𝐎𝐎!(𝐚𝐪) [𝐍𝐇𝟒

! 𝐚𝐪 ]𝐂𝟔𝐇𝟓𝐂𝐎𝐎𝐇(𝐚𝐪) [𝐍𝐇𝟑 𝐚𝐪 ]

What is the value of the equilibrium constant for the reaction of benzoic acid with ammonia? Hint: multiply the above expression by [H+]/[H+].

Multiplying the expression above by [H+] / [H+] gives:

K = 𝐂𝟔𝐇𝟓𝐂𝐎𝐎!(𝐚𝐪) [𝐍𝐇𝟒

! 𝐚𝐪 ]𝐂𝟔𝐇𝟓𝐂𝐎𝐎𝐇(𝐚𝐪) [𝐍𝐇𝟑 𝐚𝐪 ]

. [𝐇! 𝐚𝐪 ]

[𝐇! 𝐚𝐪 ]

= [𝐇! 𝐚𝐪 ] 𝐂𝟔𝐇𝟓𝐂𝐎𝐎!(𝐚𝐪)𝐂𝟔𝐇𝟓𝐂𝐎𝐎𝐇(𝐚𝐪)

. [𝐍𝐇𝟒! 𝐚𝐪 ][𝐍𝐇𝟑 𝐚𝐪 ][𝐇! 𝐚𝐪 ]

= 𝑲𝐚 × 𝑲𝒃[𝐇! 𝐚𝐪 ][𝐎𝐇! 𝐚𝐪 ]

= 𝑲𝐚 × 𝑲𝒃𝑲𝐰

= (𝟏𝟎!𝟒.𝟐𝟎) × 𝟏𝟎!𝟒.𝟕𝟔

(𝟏𝟎!𝟏𝟒) = 1.1 × 105

Answer: 1.1 × 105

What are the major species in the solution that results from adding together equal amounts of solutions A and B?

The equilibrium strong favours products so the major species are:

C6H5CO2–(aq), NH4

+(aq), H2O(l)


• Hydrochloric acid in a healthy human stomach leads to a pH in the range 1-2. What is the concentration of hydrochloric acid in the stomach?

Marks8

As pH = -log10[H3O+(aq)] and HCl is a strong acid, these pH values correspond to:

[H3O+(aq)] = 10-pH = 0.1 M at pH = 1

[H3O+(aq)] = 10-pH = 0.01 M at pH = 2

Answer: 0.01 – 0.1 M

The stomach also contains considerable amounts of chloride ions, the conjugate base of hydrochloric acid, in the form of dissolved KCl and NaCl. Is this solution a buffer? Explain your answer.

No. A buffer contains a mixture of a weak acid and its conjugate base. HCl is a very, very strong acid and so Cl- is a very, very weak base. If acid is added to the stomach, there is no base for it to react with and so the pH will lower considerably. If base is added to the stomach, it will react with the H3O+ present and the pH will rise considerably. This is not the action of a buffer.

Milk of magnesia is often taken to reduce the discomfort of acid stomach. A teaspoon of milk of magnesia, containing 0.400 g of Mg(OH)2, is added to a 0.300 L stomach solution with a pH of 1.3. What is the final pH of the solution?

The molar mass of Mg(OH)2 is (24.31 (Mg) + 2 × 16.00 (O) + 2 × 1.008 (H)) g mol-1 = 58.326 g mol-1. The number of moles in 0.400 g is therefore: number of moles = mass / molar mass = (0.400 g) / (58.326 g mol-1) = 0.00686 mol

The number of moles of OH- added to the stomach is therefore: number of moles of OH- = 2 × 0.00686 mol = 0.0137 mol



If the pH is originally 1.3, [H3O+(aq)] = 10-pH = 10-1.3 = 0.050 M. The number of moles originally present in 0.300 L is therefore: number of moles of H3O+ = concentration × volume = (0.050 mol L-1) × (0.300 L) = 0.0150 mol

This will react with the added OH-, leaving: number of moles of H3O+ = (0.0150 – 0.0137) mol = 0.0013 mol Hence, the final concentration of H3O+ is [H3O+(aq)] = number of moles / volume = (0.0013 mol) / (0.300 L) = 0.0043 M

Finally, pH = -log10[H3O+(aq)] = 2.4

Answer: 2.4


• You have completed a number of acid/base titrations during your laboratory work. What is the difference between the ‘end point’ and the ‘equivalence point’ in an acid/base titration?

Marks6

The endpoint is where the indicator changes colour and the reaction is observed to be completed. The equivalence point is where equal amounts of acid and base have been added. Ideally, the endpoint and equivalence point should be as close to one another as possible.

How do you determine the concentration of a weak acid through titration with a strong base? Include all necessary steps in your explanation.

Place a known volume (eg 25.00 mL) of the weak acid solution in a conical flask and add 2 drops of a suitable indicator (such as phenolphthalein).

Titrate with a known concentration of the strong base from a burette until the end point (the first permanent pink colour) is reached. Record the volume used.

The equation of the reaction must be known

HnA(aq) + nOH–(aq) →→→→ An–(aq) + nH2O

This can then be used to calculate the moles of OH– used.

Hence calculate the moles of weak acid present in 25.00 mL.

Hence calculate the moles of weak acid present in 1000.00 mL (i.e. its concentration).

How do you determine the pKa of a weak acid through titration with a strong base? Include all necessary steps in your explanation.

The titration described above is used and the titration of the weak acid is continued past its equivalence point with a strong base, recording the pH as you go.

Construct a graph of mL base added vs pH. This is a titration curve.

Determine the volume of base required to reach equivalence point (where slope of line is vertical).

Divide this value by 2 to give the half equivalence point (where [HA] = [A –]).

Read the value of the pH at the half equivalence point from the titration curve. From Henderson-Hasselbalch equation, this is the point where pH = pKa.


• Calculate the pH of a 0.020 M solution of Ba(OH)2. Marks

1

Ba(OH)2 is a strong base so it will completely dissociate in solution:

Ba(OH)2(s) à Ba2+(aq) + 2OH-(aq)

As each Ba(OH)2 dissociates to make 2OH-, a 0.020 M solution has [OH-(aq)] = 0.040 M. By definition, pOH = -log10[OH-(aq)] so pOH = -log10(0.040) = 1.40. As pH + pOH = 14.00,

pH = 14.00 – 1.40 = 12.60

pH = 12.60

• Calculate the pH of a 0.150 M solution of HNO2. The pKa of HNO2 is 3.15. 3

As HNO2 is a weak acid, [H3O+] must be calculated using a reaction table:

HNO2 H2O H3O+ NO2-





Ka = 𝐇𝟑𝐎! [𝐍𝐎𝟐

!][𝐇𝐍𝐎𝟐]

= 𝒙𝟐

𝟎.𝟏𝟓𝟎!𝒙


x2 = 0.150 × 10–3.15 or x = 1.03 × 10-2 M = [H3O+]


pH = −log10[H3O+] = −log10(1.03 × 10-2) = 1.987

pH = 1.987



• Calculate the pH of a solution that is 0.080 M in acetic acid and 0.160 M in sodium acetate. The pKa of acetic acid is 4.76.

2

The solution contains both a weak acid (acetic acid) and its conjugate base (the acetate ion). This is a buffer and its pH can be calculate using the Henderson-Hasselbalch equation. With [acid] = 0.080 M and [base] = 0.160 M,

pH = pKa + log[𝐛𝐚𝐬𝐞][𝐚𝐜𝐢𝐝]

= 4.76 + log𝟎.𝟏𝟔𝟎𝟎.𝟎𝟖𝟎

= 5.06

pH = 5.06


• Solution A consists of a 0.50 M aqueous solution of HF at 25 °C. Calculate the pH of Solution A. The pKa of HF is 3.17.

Marks 8

As HF is a weak acid, [H3O+] must be calculated using a reaction

table:

HF H2O H3O+ HF





23

a

[H O ][F ] xK

[HF] 0.50 x

+ −+ −+ −+ −

= == == == =−−−−

As pKa = -log10Ka, Ka = 10−−−−3.17 and is very small, 0.50 – x ~ 0.50 and hence:

x2 = 0.50 ×××× 10–3.17 or x = 1.84 × 10-2 M = [H 3O+]


pH = −−−−log10[H3O+] = −−−−log10[(1.84 × 10-2)] = 1.74

pH = 1.74



At 25 °C, 1.00 L of Solution B consists of 12.97 g of lithium fluoride, LiF, dissolved in water. Calculate the pH of Solution B.

The molar mass of LiF is 6.941 (Li) + 19.00 (F) = 25.941. Hence, the number of moles in 12.97 g is:

number of moles = mass 12.97

0.5000molarmass 25.941

= = mol

As this is dissolved in 1.00 L, [F-] = 0.500 M. F- is a weak base so [H3O

+] must be calculated via the calculation of [OH-] from the reaction table:

F- H2O HF OH–



final 0.500 - x large x x


2

b

[HF][OH ] xK

0.500 x[F ]

−−−−

−−−−= == == == =−−−−

As pKa + pKb = 14.00, pKb = 14.00 – 3.17 = 10..83. As pKb = -log10Kb, Kb = 10-10.83 and is very small. Hence, 0.500 – x ~ 0.500 and hence:

x2 = 0.500 ×××× 10–10.83 or x = 2.72 × 10-6 M = [OH -]

Hence, the pOH is given by:

pOH = −−−−log10[OH -] = −−−−log10[(2.72 × 10-6)] = 5.57

Finally, as pH + pOH = 14.00, pH = 14.00 – 5.57 = 8.43

pH = 8.43



Solution B (1.00 L) is poured into Solution A (1.00 L) and allowed to equilibrate at 25 °C. Calculate the pH of the final solution.

The solution consists of a weak acid (HF) and its conjugate base (F-). The Henderson -Hasselbalch equation can be used:

pH = pKa + 10

[base]log

[acid]

[base] = [F-] = 0.500 M and [acid] = [HF] = 0.50 M. Hence,

pH = pKa + 10

[base]log

[acid]

= 3.17 + 10

0.500log

0.50

= 3.17

pH = 3.17

If you wanted to adjust the pH of the mixture of Solution A and Solution B to be exactly equal to 4.00, which component in the mixture would you need to increase in concentration?

pH needs to increase: increase [base] (i.e. [LiF])


• What is the pH of a 0.020 M solution of HF? The pKa of HF is 3.17. Marks

2

As HF is a weak acid, [H3O+] must be calculated using a reaction table:

HF H2O H3O+ F–



final 0.020 – x large x x


23

a

[H O ][F ] xK

[HF] 0.020 x

+ −+ −+ −+ −

= == == == =−−−−

As pKa = -log10(Ka), Ka = 10−−−−3.17 and is very small. Hence, 0.020 – x ~ 0.020 and hence:

x2 = 0.020 ×××× 10–3.17 or x = 3.68 × 10-3 M = [H 3O+]


pH = −−−−log10[H3O+] = −−−−log10[(3.68 × 10-3)] = 2.43

pH = 2.43

• What is the pH of a solution that is 0.075 M in acetic acid and 0.150 M in sodium acetate? The pKa of CH3COOH is 4.76.

2

This solution containing a weak acid (CH3COOH) and its conjugate base (CH3COO-) is a buffer and the Henderson-Hasselbalch equation can be used to calculate the pH:

pH = pKa + 10

[base]log

[acid]

= pKa(CH3COOH) + 310

3

[CH COO ]log

[CH COOH]

−−−−

pH = 4.76 + 10

0.150log

0.075

= 5.06

pH = 5.06

THE ANSWER CONTINUES ON THE NEXT PAGE


• What is the pH of a 0.010 M solution of Ba(OH)2? 2

Ba(OH)2 is a strong base and dissociates completely according to the equation:

Ba(OH)2(aq) �� Ba2+(aq) + 2OH-(aq)

Hence a 0.010 M solution has [OH-(aq)] = 2 × 0.010 = 0.020 M.

As pOH = -log10([OH -(aq)], pOH = -log10(0.020) = 1.70.

In aqueous solution, pH + pOH = 14.00 so pH = (14.00 – 1.70) = 12.30

pH = 12.30


What is the pH of a 0.010 M solution of Ca(OH)2? Marks

2

Ca(OH)2 is a strong base and dissociates completely according to the equation:

Ca(OH)2(aq) Ca2+

(aq) + 2OH-(aq)

A 0.010 M solution therefore has [OH-(aq)] = 0.020 M and

pOH = -log10([OH-(aq)]) = -log10(0.020) = 1.70

As pH + pOH = 14.00, pH = (14.00 – 1.70) = 12.30

pH = 12.30

What is the pH of a 0.010 M solution of HNO2? The pKa of HNO2 is 3.15. 2

Nitrous acid is a weak acid so [H3O+] must again be calculated:

HNO2(aq) H2O(l) H3O+(aq)

NO2

–(aq)



final 0.010 – x large x x


2

3 2

a

2

[H O ][NO ]

[HNO ] 0.010

xK

x

As Ka = 103.15

is very small, 0.010 – x ~ 0.010 and hence:

x2 = 0.010 10

–3.15 or x = 2.66 × 10

-3 M = [H3O

+(aq)]


pH = log10[H3O+(aq)] = log10[(2.66 × 10

-3)] = 2.58

pH = 2.58



What is the pH of a solution that is 0.020 M in CH3COOH and 0.010 M in CH3CO2–?

The Ka of CH3COOH is 1.8 10–5

M.

2

The solution contains a mixture of a weak acid (CH3COOH) and its conjugate

base (CH3COO-) so acts as a buffer and the Henderson-Hasselbalch equation

can be used:

pH = pKa +

]acid[

]base[log10

with [base] = [CH3COO-(aq)] and [acid] = [CH3COOH(aq)].

As Ka = 1.8 × 10-5

, pKa = -log10Ka = -log10(1.8 × 10-5

) = 4.74.

Hence:

pH = 4.74 + 10

[0.010]log

[0.020]

= 4.44

pH = 4.44


• Solution A consists of a 0.15 M aqueous solution of nitrous acid (HNO2) at 25 °C. Calculate the pH of Solution A. The pKa of HNO2 is 3.15.

Marks 8

Nitrous acid is a weak acid so [H3O+] must be calculated:

HNO2 H2O H3O+ NO2–

initial 0.15 large 0 0 change -x negligible +x +x final 0.15 – x large x x


23 2

a2

[H O ][NO ][HNO ] 0.15

+ −

= =−

xKx

As Ka = 10−3.15 is very small, 0.15 – x ~ 0.15 and hence:

x2 = 0.15 × 10–3.15 or x = 1.03 × 10-2 M = [H3O+] Hence, the pH is given by:

pH = −log10[H3O+(aq)] = −log10[(1.03 × 10-2)] = 1.99

pH = 1.99



At 25 °C, 1.00 L of Solution B consists of 13.8 g of sodium nitrite (NaNO2) dissolved in water. Calculate the pH of Solution B.

The formula mass of NaNO2 is (22.99 (Na) + 14.01 (N) + 2 × 16.00 (O)) = 69. Therefore, 13.8 g corresponds to:

number of moles of NaNO2 = mass 13.8 0.200molformulamass 69.0

= =

As this is dissolved in 1.00 L, the concentration is 0.200 M. NO2

– is a weak base so [OH–(aq)] must be calculated from the equilibrium:

NO2– H2O OH– HNO2

initial 0.200 large 0 0 change -y negligible +y +y final 0.200 – y large y y


2

2a

2

[OH ][HNO ]0.2[NO ]

−

−= =

−

yKy

For an acid and its conjugate base, pKa + pKb = 14.00 so:

pKb = 14.00 – 3.15 = 10.85

As pKb = 10.85, Kb = 10−10.85. Kb is very small so 0.200 – y ~ 0.200 and hence:

y2 = 0.200 × 10–10.85 or y = 1.68 × 10-6 M = [OH−(aq)]

Hence, the pOH is given by pOH = −log10[OH−(aq)] = −log10[1.68 × 10-6] = 5.77

Finally, pH + pOH = 14 so pH = (14.00 – 5.77) = 8.23

pH = 8.23



Solution B (1.00 L) is poured into Solution A (1.00 L) and allowed to equilibrate at 25 °C. Calculate the pH of the final solution.

Solution A, HNO2(aq), is 0.15 M. When 1.00 L of A is added to 1.00 L of B, a 2.00 L solution is formed. This dilution halves the concentration to 0.075 M. Similarly, solution B, KNO2(aq), which is initially 0.200 M is diluted to 0.100 M by the addition of solution A. The combined solution contains a mixture of a weak acid (HNO2) and its conjugate base (NO2

-) so acts as a buffer and the Henderson-Hasselbalch equation can be used:

pH = pKa + ⎟⎟⎠

⎞⎜⎜⎝

⎛

]acid[]base[log10 with [base] = [NO2

-(aq)] and [acid] = [HNO2(aq)].

As pKa for HNO2 = 3.15, [NO2

-(aq)] = 1.00 M and [HNO2(aq)] = 0.075 M:

pH = 3.15 + log10[0.100][0.075]

!

"#

$

%& = 3.27

pH = 3.27

If you wanted to adjust the pH of the mixture of Solution A and Solution B to be exactly equal to 3.00, which component in the mixture would you need to increase in concentration?

acid (HNO2)

scilearn.sydney.edu.au · chem1002 2014-n-2 november 2014 • calculate the ph of a 0.020 m...

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