Question

A buffer solution contains 0.496 M hydrocyanic acid and 0.399 M sodium cyanide . If 0.0461 moles of sodium hydroxide are added to 225 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume does not change upon adding sodium hydroxide. )

Answer 1

Answer : The pH of the solution is, 9.63

Explanation : Given,

The dissociation constant for HCN = $pK_a=9.31$

First we have to calculate the moles of HCN and NaCN.

$\text{Moles of HCN}=\text{Concentration of HCN}\times \text{Volume of solution}=0.496M\times 0.225L=0.1116mole$

and,

$\text{Moles of NaCN}=\text{Concentration of NaCN}\times \text{Volume of solution}=0.399M\times 0.225L=0.08978mole$

The balanced chemical reaction is:

                          $HCN+NaOH\rightarrow NaCN+H_2O$

Initial moles     0.1116       0.0461     0.08978

At eqm.       (0.1116-0.0461)    0       (0.08978+0.0461)

                        0.0655                       0.1359

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

Now put all the given values in this expression, we get:

$pH=9.31+\log (\frac{0.1359}{0.0655})$

$pH=9.63$

Therefore, the pH of the solution is, 9.63