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Acids Lesson 10 Calculating Ka From pH

Acids Lesson 10 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. [H + ] = 0.05248 M. pH = 1.28. [H + ] = 10 -1.28. H 2 C 2 O 4 ⇄ H + + HC 2 O 4 -. I. 0.100. 0 0. C. - 0.05248. 0.05248. 0.05248. E.

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Acids Lesson 10 Calculating Ka From pH

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  1. Acids Lesson 10 Calculating Ka From pH

  2. 1. The pH of 0.100 M H2C2O4is 1.28. Calculate the Ka for the weak acid. [H+] = 0.05248 M pH = 1.28 [H+] = 10-1.28 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 0.04752 [H2C2O4]

  3. 2. If the pH of 0.40 M NH3@ 25 oC is 11.427, calculate the Kb. pH = 11.427 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 0.0026730.002673 [NH4+][OH-] (0.002673)2 = 1.8 x 10-5 Kb = = 0.3973 [NH3]

  4. 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

  5. 4. The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-. CN-+ H2O ⇄HCN + OH- I0.20 0 0 C0.002858 0.002858 0.002858 E0.19710.002858 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

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

  7. Calculate the pHof a solution made by mixing 100.0 mL of0.050 M HCl with 100 mL of water. HCl → H+ + Cl- 1( ) 2 0.050 M 0.025 M 0.025 M pH = 1.60

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