Acids Lesson 12 Calculating Ka From pH
Acids Lesson 12 Calculating Ka From pH
Jan 06, 2016
Acids Lesson 12 Calculating Ka From pH. 1.The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. 1.The pH of 0.100 M H 2 C 2 O 4 is 1.28. Calculate the Ka for the weak acid. H 2 C 2 O 4 ⇄ H + +HC 2 O 4 - I 0.100 00 C E. - PowerPoint PPT Presentation
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AcidsLesson 12
Calculating KaFrom pH
1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the
weak acid.
1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the
weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 0 0CE
1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the
weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 0 0CE
pH = 1.28
1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the
weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 0 0CE
pH = 1.28
[H+] = 10-1.28
1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the
weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 0 0CE
pH = 1.28
[H+] = 10-1.28
[H+] = 0.05248 M
1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the
weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 0 0CE 0.05248 0.05248
pH = 1.28
[H+] = 10-1.28
[H+] = 0.05248 M
1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the
weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 0 0C - 0.05248 0.05248 0.05248E 0.04752 0.05248 0.05248
pH = 1.28
[H+] = 10-1.28
[H+] = 0.05248 M
1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the
weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 0 0C - 0.05248 0.05248 0.05248E 0.04752 0.05248 0.05248
pH = 1.28
[H+] = 10-1.28
[H+] = 0.05248 M
Ka = [H+][HC2O4-]
[H2C2O4]
1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the
weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 0 0C - 0.05248 0.05248 0.05248E 0.04752 0.05248 0.05248
pH = 1.28
[H+] = 10-1.28
[H+] = 0.05248 M
Ka = [H+][HC2O4-]
= (0.05248)2
[H2C2O4] 0.04752
1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the
weak acid.
H2C2O4 ⇄ H+ + HC2O4-
I 0.100 0 0C - 0.05248 0.05248 0.05248E 0.04752 0.05248 0.05248
pH = 1.28
[H+] = 10-1.28
[H+] = 0.05248 M
Ka = [H+][HC2O4-]
= (0.05248)2 = 5.8 x 10-2
[H2C2O4] 0.04752
2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb.
2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the
Kb. NH3 + H2O ⇄ NH4
+ + OH-
I 0.40 0 0CE
2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the Kb.
NH3 + H2O ⇄ NH4+ + OH-
I 0.40 0 0CE
pH = 11.427
2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the
Kb. NH3 + H2O ⇄ NH4
+ + OH-
I 0.40 0 0CE
pH = 11.427pOH = 2.573
2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the
Kb. NH3 + H2O ⇄ NH4
+ + OH-
I 0.40 0 0CE
pH = 11.427pOH = 2.573[OH-] = 10-2.573
2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the
Kb. NH3 + H2O ⇄ NH4
+ + OH-
I 0.40 0 0CE
pH = 11.427pOH = 2.573[OH-] = 10-2.573
[OH-] = 0.002673 M
2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the
Kb. NH3 + H2O ⇄ NH4
+ + OH-
I 0.40 0 0CE 0.002673 0.002673
pH = 11.427pOH = 2.573[OH-] = 10-2.573
[H+] = 0.002673 M
2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the
Kb. NH3 + H2O ⇄ NH4
+ + OH-
I 0.40 0 0C - 0.002673 0.002673 0.002673E 0.3973 0.002673 0.002673
pH = 11.427pOH = 2.573[OH-] = 10-2.573
[OH-] = 0.002673 M
2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the
Kb. NH3 + H2O ⇄ NH4
+ + OH-
I 0.40 0 0C - 0.002673 0.002673 0.002673E 0.3973 0.002673 0.002673
pH = 11.427pOH = 2.573[OH-] = 10-2.573
[OH-] = 0.002673 M
Kb = [NH4+][OH-]
[NH3]
2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the
Kb. NH3 + H2O ⇄ NH4
+ + OH-
I 0.40 0 0C - 0.002673 0.002673 0.002673E 0.3973 0.002673 0.002673
pH = 11.427pOH = 2.573[OH-] = 10-2.573
[OH-] = 0.002673 M
Kb = [NH4+][OH-]
= (0.002673)2
[NH3] 0.3973
2. If the pH of 0.40 M NH3 @ 25 oC is 11.427, calculate the
Kb. NH3 + H2O ⇄ NH4
+ + OH-
I 0.40 0 0C - 0.002673 0.002673 0.002673E 0.3973 0.002673 0.002673
pH = 11.427pOH = 2.573[OH-] = 10-2.573
[OH-] = 0.002673 M
Kb = [NH4+][OH-]
= (0.002673)2 = 1.8 x 10-5
[NH3] 0.3973
3. The pH of a 1.0 M triprotic weak acid is 4.568. Calculate the Ka and identify the acid.
Ka = 7.3 x 10-10
Boric acid
4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
CN- + H2O ⇄ HCN + OH-
4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
CN- + H2O ⇄ HCN + OH-
I 0.20 0 0
4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
CN- + H2O ⇄ HCN + OH-
I 0.20 0 0
C 0.002858 0.002858 0.002858
4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
CN- + H2O ⇄ HCN + OH-
I 0.20 0 0
C 0.002858 0.002858 0.002858
E 0.1971 0.002858 0.002858
4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
CN- + H2O ⇄ HCN + OH-
I 0.20 0 0
C 0.002858 0.002858 0.002858
E 0.1971 0.002858 0.002858
[OH-] = 0.002858 M
4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
CN- + H2O ⇄ HCN + OH-
I 0.20 0 0
C 0.002858 0.002858 0.002858
E 0.1971 0.002858 0.002858
[OH-] = 0.002858 M
Kb = [HCN][OH-]=
[CN-]
4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.
CN- + H2O ⇄ HCN + OH-
I 0.20 0 0
C 0.002858 0.002858 0.002858
E 0.1971 0.002858 0.002858
[OH-] = 0.002858 M
Kb = [HCN][OH-]= (0.002858)2 = 4.1 x 10-5
[CN-] 0.1971
5. Calculate the pH of 0.020 M H3BO3
H3BO3 ⇄ H+ + H2BO3-
I 0.020 M 0 0C x x xE 0.020 - x x x
0 small ka
x2 = 3.8 x 10-10
0.020
x = [H+] = 2.76 x 10-6 M
pH = -Log[2.76 x 10-6]
pH = 5.42
2 sig figs due to molarity and Ka
6. Calculate the pH of a solution made by mixing 100.0 mL of 0.050 M HCl with 100 mL of water.
HCl → H+ + Cl-
1(0.050 M) 0.025 M 0.025 M2
pH = 1.60
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