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}* \text{Volume of solution}=0.496M* 0.225L=0.1116mole`

and,

`\text{Moles of NaCN}=\text{Concentration of NaCN}* \text{Volume of solution}=0.399M* 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 ([Salt])/([Acid])`

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

`pH=9.31+\log ((0.1359)/(0.0655))`

`pH=9.63`

Therefore, the pH of the solution is, 9.63