Respuesta :
Explanation:
The given reaction equation is as follows.
[tex]HCN + NaOH \rightarrow NaCN + H_{2}O[/tex]
Hence, initial moles of HCN will be as follows.
Moles = Molarity × Volume
= [tex]0.4006 M \times 120 ml[/tex]
= 48.072 mol
Now, we will calculate the volume of NaOH as follows.
[tex]M_{1}V_{1} = M_{2}V_{2}[/tex]
[tex]0.4006 \times 120 = 0.6812 \times V_{2}[/tex]
[tex]V_{2}[/tex] = 70.57 ml
At the equivalence point, moles of both HCN and NaOH will be equal. And, total volume will be as follows.
120 ml + 70.57 ml = 190.57 ml
Initial concentration of [tex]CN^{-}[/tex] is as follows.
[tex]\frac{48.072}{190.57}[/tex]
= 0.25 M
Now, the equilibrium equation will be as follows.
[tex]CN^{-} + H_{2}O \rightleftharpoons HCN + OH^{-}[/tex]
Initial: 0.25
Equilbm: 0.25 - x x x
Expression to find [tex]K_{a}[/tex] is as follows.
[tex]K_{a} = \frac{x^{2}}{0.25 - x}[/tex]
We know that, [tex]pK_{a} = -log K_{a}[/tex]
[tex]K_{a}[/tex] = antilog (-9.21)
= [tex]6.16 \times 10^{-10}[/tex]
As, [tex]K_{b} = \frac{K_{w}}{K_{a}}[/tex]
[tex]\frac{10^{-14}}{6.16 \times 10^{-10}} = \frac{x^{2}}{0.25 - x}[/tex]
x = [tex]0.2 \times 10^{-2} = [OH^{-}][/tex]
Now, we will calculate the pH as follows.
pOH = [tex]-log[OH^{-}][/tex]
as pOH = 2.698
And, pH + pOH = 14
pH = 14 - 2.698
= 11.30
Thus, we can conclude that pH at equivalence is 11.30.
