logarithms, ph, poh, pk a. logarithms logarithmpower to which you must raise a base number to obtain...

Logarithms, pH, pOH, pKa

LOGARITHMSLOGARITHM power to which you must raise a

base number to obtain the desired number

Log (base 10)

• Logs (base 10) = exponent to which 10 must be raised to obtain desired number

Log (base 10)

• Logs (base 10) = exponent to which 10 must be raised to obtain desired number

• log 1 = 0 because 1 = 100

Log (base 10)

• Logs (base 10) = exponent to which 10 must be raised to obtain desired number

• log 1 = 0 because 1 = 100

• Common log of a number that is a power of 10 is always a whole number

Log (base 10)

•log is exponent of 10 WITH ITS SIGN eg -4 log 10-4 = -4

• Fractional power of 10 on calculator, enter number and use log button,

log (2 x 10-4) = -3.7

Inverse log (base 10)

• Inverse log used to find number corresponding to log = 10x

Inverse log (base 10)• Inverse log used to find number corresponding to log = 10x

• Whole integer log: use log (given number) with its sign as power of 10

Inverse log (base 10)• Inverse log used to find number corresponding to log = 10x

• Whole integer log: use log (given number) with its sign as power of 10

• Decimal log: enter number WITH ITS SIGN Use the:

Inverse log (base 10)• Inverse log used to find number corresponding to log = 10x

• Whole integer log: use log (given number) with its sign as power of 10

• Decimal log: enter number WITH ITS SIGN and use either

• 10x button OR

Inverse log (base 10)• Inverse log used to find number corresponding to log = 10x

• Whole integer log: use log (given number) with its sign as power of 10

• Decimal log: enter number WITH ITS SIGN and use either

• 10x button OR

• [2nd][log] buttons

Example #1 – inverse logWhat number corresponds to a

log= -5.34?

Example #1 – inverse logWhat number corresponds to a

log= -5.34?

Inverse log –5.34 = 10-5.34

Example #1 Solution

[2nd]log –5.34 = 10-5.34 = 4.6 x 10-6

pH

pH

pH A measure of the acidity of aqueous solutions

pH

pH A measure of the acidity of aqueous solutions

pH = - log [H3O+]

pH

pH A measure of the acidity of aqueous solutions

pH = - log [H3O+]

pOH = - log [OH-]

pH

pH A measure of the acidity of aqueous solutions

pH = - log [H3O+]

pOH = - log [OH-]

pK = - log [Keq]

pH

pH A measure of the acidity of aqueous solutions

pH = - log [H3O+]

pOH = - log [OH-]

pK = - log [Keq]

p(anything) = - log (anything)

[H3O+] from pH[OH-] from pOH

[H3O+] = 10 - pH

[H3O+] from pH[OH-] from pOH

[H3O+] = 10 - pH

[OH-] = 10 - pOH

[H3O+] from pH[OH-] from pOH

[H3O+] = 10 - pH

[OH-] = 10 - pOH

• the higher the pH, the lower the [H3O+] in the solution

pH in a NEUTRAL SOLUTION

neutral solution: [H3O+] = [OH-] = 1.0 x 10-7 M

pH in a NEUTRAL SOLUTION

neutral solution: [H3O+] = [OH-] = 1.0 x 10-7 M

pH = pOH = 7.00

SUMMARY OF pH

pH pOH

acidic < 7

SUMMARY OF pH

pH pOH

acidic < 7

neutral 7

SUMMARY OF pH

pH pOH

acidic < 7

neutral 7

basic > 7

SUMMARY OF pH

pH pOH

acidic < 7 > 7

neutral 7

basic > 7

SUMMARY OF pH

pH pOH

acidic < 7 > 7

neutral 7 7

basic > 7

SUMMARY OF pH

pH pOH

acidic < 7 > 7

neutral 7 7

basic > 7 < 7

pH CHANGE

• change of 1 pH unit means [H3O+] has undergone a 10-fold change

pH CHANGE

• change of 1 pH unit means [H3O+] has undergone a 10-fold change

• pH 5 vs pH 6:

pH CHANGE

• change of 1 pH unit means [H3O+] has undergone a 10-fold change

• pH 5 vs pH 6:

• change of one pH unit; change [ ] by factor of 10

pH CHANGE

• change of 1 pH unit means [H3O+] has undergone a 10-fold change

• pH 5 vs pH 6:

• change of one pH unit; change [ ] by factor of 10

• pH 5 10X more acidic than pH 6

pH CHANGE

• change of 1 pH unit means [H3O+] has undergone a 10-fold change

• pH 5 vs pH 6:

• change of one pH unit; change [ ] by factor of 10

• pH 5 10X more acidic than pH 6

• pH 4 100X more acidic than pH 6

pKW

Kw = [H3O+][OH-] = 1.0 x 10-14

pKW

Kw = [H3O+][OH-] = 1.0 x 10-14

pH + pOH = 14.00 = pKw

pH of STRONG ACID

Example #2 Calculations Involving pH and pOH

Calculate [H3O+], pH, [OH-], and pOH for a 0.015 M HNO3 solution.

Example #2 Solution

HNO3(aq) + H2O H3O+(aq) + NO3-(aq)

0.015 M 0.015 M

Example #2 Solution

HNO3(aq) + H2O H3O+(aq) + NO3-(aq)

0.015 M 0.015 M

pH = - log [H3O+]

Example #2 Solution

HNO3(aq) + H2O H3O+(aq) + NO3-(aq)

0.015 M 0.015 M

pH = - log [H3O+] = - log 0.015 = - (-1.82) = 1.82

Example #2 Solution

pH + pOH = 14.00

Example #2 Solution

pH + pOH = 14.00

pOH = 14.00 - pH

Example #2 Solution

pH + pOH = 14.00

pOH = 14.00 - pH = 14.00 - 1.82 = 12.18

Example #2 Solution

pH + pOH = 14.00

pOH = 14.00 - pH = 14.00 - 1.82 = 12.18

pOH = - log [OH-]

Example #2 Solution

pH + pOH = 14.00

pOH = 14.00 - pH = 14.00 - 1.82 = 12.18

pOH = - log [OH-]

12.18 = - log [OH-]

Example #2 Solution

pH + pOH = 14.00

pOH = 14.00 - pH = 14.00 - 1.82 = 12.18

pOH = - log [OH-]

12.18 = - log [OH-]

[OH-] = [2nd ]log -12.18

Example #2 Solution

pH + pOH = 14.00

pOH = 14.00 - pH = 14.00 - 1.82 = 12.18

pOH = - log [OH-]

12.18 = - log [OH-]

[OH-] = [2nd] log -12.18 = 10-12.18

Example #2 Solution

pH + pOH = 14.00

pOH = 14.00 - pH = 14.00 - 1.82 = 12.18

pOH = - log [OH-]

12.18 = - log [OH-]

[OH-] = [2nd]log -12.18 =10-12.18 = 6.7 x 10-13 M

pH of STRONG BASE

Example #3 Calculations Involving pH and pOH

Calculate [H3O+], pH, [OH-], and pOH for a 0.015 M Ca(OH)2 solution.

Example #3 Solution

Ca(OH)2(aq) Ca2+(aq) + 2 OH-(aq)

Example #3 Solution

Ca(OH)2(aq) Ca2+(aq) + 2 OH-(aq) 0.015 M 2 x 0.015 M = 0.030 M

Example #3 Solution

Ca(OH)2(aq) Ca2+(aq) + 2 OH-(aq) 0.015 M 2 x 0.015 M = 0.030 M

pOH = - log [OH-]

Example #3 Solution

Ca(OH)2(aq) Ca2+(aq) + 2 OH-(aq) 0.015 M 2 x 0.015 M = 0.030 M

pOH = - log [OH-] = - log 0.030 = - (-1.52) = 1.52

Example #3 Solution

pH + pOH = 14.00

Example #3 Solution

pH + pOH = 14.00

pH = 14.00 - pOH

Example #3 Solution

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 1.52 = 12.48

Example #3 Solution

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 1.52 = 12.48

pH = - log [H3O+]

Example #3 Solution

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 1.52 = 12.48

pH = - log [H3O+]

12.48 = - log [[H3O+]

Example #3 Solution

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 1.52 = 12.48

pH = - log [H3O+]

12.48 = - log [[H3O+]

[H3O+] = [2nd]log -12.48

Example #3 Solution

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 1.52 = 12.48

pH = - log [H3O+]

12.48 = - log [[H3O+]

[H3O+] = [2nd]log -12.48 = 10-12.48

Example #3 Solution

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 1.52 = 12.48

pH = - log [H3O+]

12.48 = - log [[H3O+]

[H3O+] = [2nd]log -12.48 =10-12.48 =3.3 x 10-13 M

pKa

• pKa = - log Ka

pKa

• pKa = - log Ka

•The lower the pKa, the stronger the acid

pKa

• pKa = - log Ka

•The lower the pKa, the stronger the acid

• Used with weak acids• Strong acid have Ka approaching infinity

REVIEW

• Auto-ionization of H2O: Kw = [H3O+][OH-]

• pH = - log [H3O+] = - log [H+]

• pOH = - log [OH-]

• pK = - log K