Acids Lesson 10 Calculating Ka From pH

1. The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the weak acid. [H+] = 0.05248 M H2C2O4 ⇄ H+ + HC2O4- I 0.100 0 0 C - 0.05248 0.05248 0.05248 E 0.04752 0.05248 0.05248 [H+][HC2O4-] Ka = (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. pOH = 2.573 [OH-] = 10-2.573 [OH-] = 0.002673 M NH3 + H2O ⇄ NH4+ + OH- I C E 0.40 0 0 - 0.002673 0.002673 0.002673 0.3973 0.002673 0.002673 [NH4+][OH-] (0.002673)2 = 1.8 x 10-5 Kb = = 0.3973 [NH3]

3. The pH of a 1. 0 M triprotic weak acid is 4. 568 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 Hint- it’s not very exciting Boric acid

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 pH = 11.456 pOH = 2.544 [OH-] = 0.002858 M [HCN][OH-] (0.002858)2 = 4.1 x 10-5 Kb = = [CN-] 0.1971

5. If the pH of H3BO3 is 5.42, calculate the initial concentration. [H+] = 3.802 x 10-6 M H3BO3 ⇄ H+ + H2BO3- x I C 3.802 x 10-6 3.802 x 10-6 3.802 x 10-6 E x - 3.802 x 10-6 3.802 x 10-6 3.802 x 10-6 0 small ka (3.802 x 10-6)2 = 7.3 x 10-10 x x = 0.020M

Calculate the pH of a solution made by mixing 100. 0 mL of 0 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( ) 2 0.050 M 0.025 M 0.025 M pH = 1.60