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}