Answer:
pH of acetic acid solution is 2.88
Explanation:
[tex]pK_{a}=4.75[/tex]
or, [tex]-log(K_{a})=4.75[/tex]
or, [tex]K_{a}=10^{-4.75}=1.78\times 10^{-5}[/tex]
We have to construct an ICE table to determine concentration of [tex]H^{+}[/tex] and corresponding pH. Initial concentration of acetic acid is 0.1 M.
[tex]CH_{3}COOH\rightleftharpoons CH_{3}COO^{-}+H^{+}[/tex]
I(M): 0.1 0 0
C(M): -x +x +x
E(M): 0.1-x x x
So, [tex]\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}=K_{a}[/tex]
or, [tex]\frac{x^{2}}{0.1-x}=1.78\times 10^{-5}[/tex]
or, [tex]x^{2}+(1.78\times 10^{-5}\times x)-(1.78\times 10^{-6})=0[/tex]
So, [tex]x=\frac{-(1.78\times 10^{-5})+\sqrt{(1.78\times 10^{-5})^{2}+(4\times 1\times 1.78\times 10^{-6})}}{2\times 1}[/tex](M)
so, [tex]x=1.33\times 10^{-3}M[/tex]
Hence [tex]pH=-log[H^{+}]=-log(1.33\times 10^{-3})=2.88[/tex]
