A chemist titrates 100.0mL of a 0.3444M hydrocyanic acid HCN solution with 0.8414M KOH solution at 25°C . Calculate the pH at equivalence. The pKa of hydrocyanic acid is 9.21 . Round your answer to 2 decimal places.

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Alleei Alleei · Answer 1 · 2020-03-31T07:13:19+02:00

Answer : The pH at equivalence is, 11.30

Explanation : Given,

Concentration of [tex]HCN[/tex] = 0.3444 M

Volume of [tex]HCN[/tex] = 100.0 mL = 0.1 L (1 L = 1000 mL)

First we have to calculate the moles of [tex]HCN[/tex]

[tex]\text{Moles of }HCN=\text{Concentration of }HCN\times \text{Volume of }HCN[/tex]

[tex]\text{Moles of }HCN=0.3444M\times 0.1L=0.03444mol[/tex]

As we known that at equivalent point, the moles of HCN and KOH are equal.

So, Moles of KOH = Moles of HCN = 0.03444 mol

Now we have to calculate the volume of KOH.

[tex]\text{Volume of }KOH=\frac{\text{Moles of }KOH}{\text{Concentration of }KOH}[/tex]

[tex]\text{Volume of }KOH=\frac{0.03444mol}{0.8414M}[/tex]

[tex]\text{Volume of }KOH=0.0409L[/tex]

Total volume of solution = 0.1 L + 0.0409 L = 0.1409 L

Now we have to calculate the concentration of KCN.

The balanced equilibrium reaction will be:

[tex]HCN+KOH\rightleftharpoons KCN+H_2O[/tex]

Moles of KCN = 0.03444 mol

[tex]\text{Concentration of }KCN=\frac{0.03444mol}{0.1409L}=0.244M[/tex]

At equivalent point,

[tex]pH=\frac{1}{2}[pK_w+pK_a+\log C][/tex]

Given:

[tex]pK_w=14\\\\pK_a=9.21\\\\C=0.244M[/tex]

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

[tex]pH=\frac{1}{2}[14+9.21+\log (0.244)][/tex]

[tex]pH=11.30[/tex]

Therefore, the pH at equivalence is, 11.30