Question

A buffer solution was created by combining 0.700 moles of HCN and 0.900 moles of NaCN in 1.00 L of water. How many moles of HCl would be added to reach a pH of 8.95?

Answer 1

Given that

[HCN] = 0.7 mol

    [NaCN] = 0.9 mol

[HCl] = ? mol

Original pH = pKa + log [NaCN]/[HCN] ----------(1)

pH after addition of HCl:

HCl + NaCN---------> HCN + NaCl

Hence,

new pH = pKa + log {[NaCN] - [HCl]/[HCN]+[HCl]} --------(1)

We know that pKa of HCN = 9.2

new pH = pKa + log {[NaCN] - [HCl]/[HCN]+[HCl]} --------(1)

8.95 = 9.2 + log {0.9 - [HCl] / 0.7 + [HCl] }

0.9 - [HCl] / 0.7 + [HCl] = 0.56

0.9 - [HCl] = 0.392 + 0.56 [HCl]

1.56 [HCl] = 0.508

[HCl ] = 0.325 moles

Therefore,

0.325 moles of HCl must be added to reach a pH of 8.95.