Answer: 2.01
Explanation:
[tex]C_6H_5COOH\rightleftharpoons C_6H_5COO^-+H^+[/tex]
initial : c 0 0
eqm: [tex]c(1-\alpha)[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
[tex]K_a=\frac{c\alpha\times c\alpha}{c(1-\alpha)}[/tex]
when [tex]\alpha[/tex] is very very small the, the expression will be,
[tex]k_a=\frac{c^2\alpha^2}{c}=c\alpha^2\\\\\alpha=\sqrt{\frac{k_a}{c}}[/tex]
And,
[tex][H^+]=c\alpha[/tex]
Thus the expression will be,
[tex][H^+]=\sqrt{k_a\times c}[/tex]
Now put all the given values in this expression, we get
[tex][H^+]=\sqrt{(6.4\times 10^{-5})\times 1.5}[/tex]
[tex][H^+]=9.7\times 10^{-3}M[/tex]
[tex]pH=-log[H^+][/tex]
[tex]pH=-log[9.7\times 10^{-3}]=2.01[/tex]
