`a)`

`HCOOH`  $\rightleftharpoons$  `H^{+}+HCOO^-`

Đặt `[H^+]=a(M)`

`->C_{HCOOH\ pu}=[HCOO^-]=a(M)`

`->[HCOOH]=0,1-a(M)`

`->K_a={a^2}/{0,1-a}=1,77.10^{-4}`

`->a≈4,12.10^{-3}`

`->pH=-log[H^+]≈2,385`

`b)`

`H_2SO_4->H^{+}+HSO_4^-`

`HSO_4^{-}`  $\rightleftharpoons$  `H^{+}+SO_4^{2-}`

Đặt `[H^+]=b(M)`

`->C_{HSO_4^-\ pu}=[SO_4^{2-}]=b(M)`

`->[HSO_4^-]=x-b(M)`

`->K_{a_2}={b^2}/{x-b}=1,2.10^{-2}`

`->b^2=0,012x-0,012b`

`->x={b^2}/{0,012}-b`

Cho cùng thể tích `1L`

`\sum n_{H^+}=4,12.10^{-3}+b(mol)`

`->[H^+\ sau]={4,12.10^{-3}+b}/2=2,06.10^{-3}+0,5b(M)`

`pH(sau)=2,385-0,385=2`

`->[H^+\ sau]=0,01=2,06.10^{-3}+0,5b`

`->b=0,01588`

`->x={0,01588^2}/{0,012}-0,01588≈5,13.10^{-3}`