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 NaOH would be added to reach a pH of 9.75?

Answer 1

Given that

[HCN] = = 0.7 mol

    [NaCN] = 0.9 mol

[NaOH] = ?

Original pH = pKa + log {[NaCN] / [HCN] }

pH after addition of NaOH:

HCN + NaOH ---------> NaCN + H2O

Hence,

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

Given that new pH = 9.75

we know that pKa of HCN = 9.1

Then, substitute all the values in Eq (1)

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

9.75 = 9.1 + log { 0.9 + [NaOH]/ 0.7-[NaOH]}    

log { 0.9 + [NaOH]/ 0.7-[NaOH]}    = 0.65

{ 0.9 + [NaOH]/ 0.7-[NaOH]}     = 4.46

0.9 + [NaOH] = 3.12 - (4.46)[NaOH]

5.46 [NaOH] = 4.02

[NaOH] = 0.736

Therefore,

0.736 moles of NaOH would be added to reach a pH of 9.75