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. )

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Alleei Alleei · Answer 1 · 2020-03-11T08:24:52+01:00

Answer : The pH of the solution is, 9.63

Explanation : Given,

The dissociation constant for HCN = [tex]pK_a=9.31[/tex]

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

[tex]\text{Moles of HCN}=\text{Concentration of HCN}\times \text{Volume of solution}=0.496M\times 0.225L=0.1116mole[/tex]

and,

[tex]\text{Moles of NaCN}=\text{Concentration of NaCN}\times \text{Volume of solution}=0.399M\times 0.225L=0.08978mole[/tex]

The balanced chemical reaction is:

[tex]HCN+NaOH\rightarrow NaCN+H_2O[/tex]

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 :

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

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

[tex]pH=9.31+\log (\frac{0.1359}{0.0655})[/tex]

[tex]pH=9.63[/tex]

Therefore, the pH of the solution is, 9.63