Question

A buffer is prepared such that [CH₃COOH]₀ = [CH₃COO^–]₀ = 0.100 M. What volume of 0.125 M NaOH (aq) can be added to 200.0 mL of this buffer solution before its buffering capacity is lost. Assume that the buffer capacity is lost when the ratio [base]₀/[acid]₀ is less than 0.1 or greater than 10.

Select one:

a. 123 mL

b. 145 mL

c. 113 mL

d. 131 mL

Question 2

Given a solution of 0.100 M benzoic acid, C₆H₅COOH (aq), and a solution of 0.150 M NaC₆H₅COO (aq), how would you prepare 100.0 mL of a buffer solution with a pH of 4.35? pK_a = 4.20 for C₆H₅COOH (aq).

Select one:

a. Mix 54.0 mL of the C₆H₅COOH (aq) with 46.0 mL of the NaC₆H₅COO (aq).

b. Mix 50.0 mL of the C₆H₅COOH (aq) with 50.0 mL of the NaC₆H₅COO (aq).

c. Mix 51.5 mL of the C₆H₅COOH (aq) with 48.5 mL of the NaC₆H₅COO (aq).

d. Mix 48.0 mL of the C₆H₅COOH (aq) with 52.0 mL of the NaC₆H₅COO (aq).

Answer 1

pH = pka + log(salt/acid)

if the [CH3COOH]0 = [CH3COO–]0 = 0.100 M

pH = pka = 4.74

by the addition of NaOH pH increases,maximum buffer capacity up to 5.74

pH = 5.74

No of mol of buffer = 0.2*0.1 = 0.02 mol

pH = pka + log(salt+x/acid-x)

5.74 = 4.74 + log((0.02+x)/(0.02-x))

x = 0.01636 mol

Volume of NaOH = n/M = 0.01636/0.125 = 0.13088 L

= 130.88 ml

answer: d.

2) pH = pka + log(salt/acid)

4.35 = 4.2+log(x)

x= 1.41

so that

x = salt/acid   = 1.41

form the options

C option is having ratio = (48.5*0.15)/(0.1*51.5) = 1.41

answer: C