Đáp án:

Ta có:

`HCl->H^{+}+Cl^{-}`

`=>[H^{+}]=[HCl]=0,001(M)`

`CH_3COOH`$\rightleftharpoons$`H^{+}+CH_3COO^{-}`

Theo đề:

`[CH_3COOH]_{"bđ"}=0,1(M)`

`[H^{+}]_{"bđ"}=0,001(M)`

Gọi:

`[CH_3COOH]_{"pl"}=[H^{+}]_{"pl"}=[CH_3COO^{-}]_{"pl"}=a(M)`

`=>[CH_3COOH]_{"cb"}=0,1-a\ (M)`

`=>[H^{+}]_{"cb"}=0,001+a\ (M)`

`=>[CH_3COO^{-}]_{"cb"}=a(M)`

`=>K_a={[H^{+}].[CH_3COO^{-}]}/{[CH_3COOH]}`

`<=>1,75.10^{-5}={(0,001+a).a}/{(0,1-a)}`

`=>a~~9,086.10^{-4}(M)`

`=>\sum[H^{+}]=0,001+9,086.10^{-4}=1,9086.10^{-3}(M)`

`=>pH=-log[H^{+}]=-log(1,9086.10^{-3})~~2,72`