What is the pH of a solution prepared by mixing 10.0 mL of 0.100 M NaOH with 15.

Question

What is the pH of a solution prepared by mixing 10.0 mL of 0.100 M NaOH with 15.0 mL of 0.100 M HCl? Assume the volumes are additive. Show how formic acid, HCOOH, dissociates in solution then calculate the pH of a 0.15 M solution of the acid. Ka for formic acid is 1.8 times 10^-4 Show how methylamine, CH_3NH_2, dissociates in solution then calculate the pH of a 0.15 M solution of the base. Kb for methylamine is 4.4 times 10^-4 Calculate K_a for the anilinium ion given that K_b for aniline (C_6H_5NH_2) is 1.5 times 10^-10 Calculate K_b for the cyanide ion given that K_a for HCN(aq) is 1.5 times 10^-10 Show how the following ions are hydrolyzed in water and tell if the ion would give an acidic or basic solution: CH_3CO_2^- NH_4^+ F^- Fe^3+ CH_3NH_3^+ Calculate the pH of a buffer solution prepared by mixing 10.0 mL of 0.25 M ammonia and 15.0 mL of 0.15 M in ammonium chloride. Assume the volumes are additive. Kb for ammonia is 1.8 times 10^-5 Calculate the pH of a buffer solution prepared by mixing 15.0 mL of 0.20 M acetic acid and 15.0 mL of 0.20 M sodium acetate. Assume the volumes are additive. Ka for acetic acid is 1.8 times 10^-5 Give a recipe to prepare an acetate acetic acid buffer having a pH of 4.90. Ka for acetic acid is 1.8 times 10^-5 Give a recipe to prepare a ammonia ammonium chloride buffer having a pH of 8.50. Kb for ammonia is 1.8 times 10-5

Explanation / Answer

1. excess [H+] = 0.1 M x 5 ml/25 ml = 0.02 M

pH = -log[H+] = 1.7

2. ionization,

HCOOH + H2O <==> HCOO- + H3O+

[H3O+] = sq.rt.(1.8 x 10^-4 x 0.15) = 5.2 x 10^-3 M

pH = -log[H3O+] = 2.28

3. ionization

CH3NH2 + H2O <==> CH3NH3+ + OH-

[OH-] = sq.rt.(4.4 x 10^-4 x 0.15) = 8.12 x 10^-3 M

pOH = -log[OH-] = 2.09

pH = 14 - pOh = 11.91

4.

a. Ka = Kw/Kb

         = 1 x 10^-14/1.5 x 10^-10 = 6.67 x 10^-5

b. Kb = Kw/Ka

         = 1 x 10^-14/1.5 x 10^-10 = 6.67 x 10^-5

5. ionization

i) CH3CO2- + H2O <==> CH3CO2H + OH-

solution : basic

ii) NH4+ + H2O <==> NH3 + H3O+

solution : acidic

iii) F- + H2O <==> HF + OH-

solution : basic

iv) Fe3+ + 3H2O <==> Fe(OH)3 + 3H+

solution : neutral (Fe(OH)3 is strong electrolyte)

v) CH3NH3+ + H2O <==> CH3NH2 + H3O+

solution : acidic

6. buffer

pH = pKa + log(base/acid)

      = 9.25 + log[(0.25 x 0.15/25)/(0.15 x 15/25)]

      = 9.296

7. pH = pKa

          = 4.745

8. To prepare buffer

take (CH3COO-/CH3COOH) in the ratio = inv.log(4.90 - 4.75) = 1.41

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