ACIDS, BASES, AND BUFFERS Calculate the pH of a 0.100 M NH_4Cl solution. Determi

Question

ACIDS, BASES, AND BUFFERS Calculate the pH of a 0.100 M NH_4Cl solution. Determine the % dissociation of the ammonium ion. What is the pH of a solution prepared by adding 50.0 mL of 0.10 M acetic acid and 30.0 mL of 0.10 M sodium acetate? What is the pH after the addition of 5.0 mL of 0.10 M HCl to the solution in 3(a)? What is the pH after the addition of 5.0 mL of 0.10 M NaOH to the solution in 3(a)? Is this a buffer solution? Explain. The pH of a 0.10 M solution of a base is found to be 11.8. What is the K_b for this base? What is the dissociation of this base? Background Information: Acids and Bases, Equilibrium Constants, Percent Ionization

Explanation / Answer

1. NH+ + H2O <==> NH3 + H3O+

Ka = 1 x 10^-14/1.8 x 10^-5 = x^2/0.1

x = [H3O+] = 7.45 x 10^-6 M

pH = -log[H3O+] = 5.13

2. % dissociation = 7.45 x 10^-6 x 100/0.1 = 0.00745%

3. (a) pH = pKa + log(sodium acetate/acetic acid)

               = 4.74 + log(0.1 x 30/0.1 x 50)

               = 4.52

(b) after 0.1 M x 5 ml = 0.5 mmol HCl added

pH = pKa + log(sodium acetate/acetic acid)

               = 4.74 + log[(0.1 x 30 - 0.5)/(0.1 x 50 + 0.5)]

               = 4.40

(c) after 0.1 M x 5 ml = 0.5 mmol NaOH added

pH = pKa + log(sodium acetate/acetic acid)

               = 4.74 + log[(0.1 x 30 + 0.5)/(0.1 x 50 - 0.5)]

               = 4.63

(d) This is a buffer solution as it reacts with added acid and base to maintain pH of solution

4. (a) pH = 11.8

pOH = 14 - pH = 2.2

pOH = -log[OH-]

[OH-] = [BH+] = 6.31 x 10^-3 M

Kb = [BH+][OH-]/[B]

     = (6.31 x 10^-3)^2/0.1

     = 3.98 x 10^-4

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