Respuesta :
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
