Question

A buffer solution contains 0.337 M KHSO3 and 0.289 M Na2SO3. Determine the pH change when 0.083 mol HBr is added to 1.00 L of the buffer.

Answer 1

this is a buffer since

HSO3- is the weak acid and SO3-2 the conjugate base

pH = pKa2 + log(SO3-2 / HSO3-)

Ka2 = 6.3*10^-8

pKa2 = -log(Ka2) = -log(6.3*10^-8) = 7.2

initial pH:

pH = pKa2 + log(SO3-2 / HSO3-)

pH = 7.2+ log(0.289/0.337) = 7.133

b)

after adding

[H+] = 0.083 mol

then,

SO3-2 left= 0.289 -0.083 = 0.206

HSO3- formed = 0.337 +0.083= 0.42

substitute

pH = pKa2 + log(SO3-2 / HSO3-)

pH = 7.2+ log(0.206 /0.42) = 6.890

pH change = 6.890 - 7.133 = -0.243 or decrease in 0.243 units