Question

Enter your answer in the provided box. How much HCI must be added to a liter of buffer that is 1.4 M in acetic acid and 0.70 M in sodium acetate to result in a buffer pH of 4.02? Imol

Answer 1

mol of HCl added = n mol

Before adding HCl

Before Reaction:

mol of CH3COO- = 0.7 M *1.0 L

mol of CH3COO- = 0.7 mol

mol of CH3COOH = 1.4 M *1.0 L

mol of CH3COOH = 1.4 mol

N mol HCl will react with n of CH3COO- to form extra n of CH3COOH

After adding HCl

mol of CH3COOH = 1.4+n mol

mol of CH3COO- = 0.7-n mol

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

use:

pH = pKa + log {[conjugate base]/[acid]}

4.02 = 4.745+log {[CH3COO-]/[CH3COOH]}

log {[CH3COO-]/[CH3COOH]} = -0.7247

[CH3COO-]/[CH3COOH] = 0.1885

So,

(0.7-n)/(1.4+n) = 0.1885

0.7-n = 0.2639 + 0.1885*n

1.1885*n = 0.4361

n = 0.367 mol

Answer: 0.367 mol