Answer:
The pH of the solution is 5.31.
Explanation:
Let "[tex]\alpha[/tex] is the dissociation of weak acid - HCN.
The dissociation reaction of HCN is as follows.
[tex]HCN+H_{2}O\rightarrow H_{3}O^{+}+CN^{-}[/tex]
Initial C 0 0
Equilibrium c(1- [tex]\alpha[/tex]) c[tex]\alpha[/tex] c[tex]\alpha[/tex]
Dissociation constant = [tex]Ka= c\alpha \times \frac{c\alpha}{c(1-\alpha)}[/tex]
[tex]=\frac{c\alpha^{2}}{(1-\alpha)}[/tex]
In this case weak acids [tex]\alpha[/tex] is very small so, (1-[tex]\alpha[/tex] ) is taken as 1.
[tex]Ka=C\alpha^{2}[/tex]
[tex]\alpha=\sqrt\frac{ka}{c}[/tex]
From the given the concentration = 0.050 M
Substitute the given value.
[tex]\alpha=\sqrt\frac{4.9\times 10^{-10}}{0.05}=9.8\times 10^{-4}[/tex]
[tex][H_{3}O^{+}]=c\alpha[/tex]
[tex][H_{3}O^{+}]=0.05\times 9.8\times 10^{-4}= 4.9\times10^{-6}[/tex]
[tex]pH= -log[H_{3}O^{+}][/tex]
[tex]=-log[4.9\times10^{-6}][/tex]
[tex]=6-log 4.9= 5.31[/tex]
Therefore, The pH of the solution is 5.31.
