Đáp án:
$$H_2SO_4\to H^++HSO_4^-$$ 
$$HSO_4^-\rightleftharpoons H^++SO_4^{2-}\quad K_a=1,2.10^{-2}$$ 
$$H_2O\rightleftharpoons H^++OH^-\quad K_n=10^{-14}$$ 
$$C_{HSO_4^-}=C_{H_2SO_4}=0,1M$$ 
Ta có: 
$$[H^+]=[OH^-]+[HSO_4^-]+2[SO_4^{2-}]$$ 
Đặt $$h=[H^+]\Rightarrow h=\dfrac{10^{-14}}{h}+[HSO_4^-]+2[SO_4^{2-}]$$ 
$$K_a=\dfrac{h[SO_4^{2-}] }{[HSO_4^{-}]}$$ 
$$\Rightarrow [HSO_4^{-}]=\dfrac{h[SO_4^{2-}]}{K_a}$$ 
$$\Rightarrow h=\dfrac{10^{-14}}{h}+\dfrac{h[SO_4^{2-}]}{K_a}+2[SO_4^{2-}]$$ 
Mà $$[HSO_4^-]=0,1-[SO_4^{2-}]=\dfrac{h[SO_4^{2-}]}{K_a}$$ 
$$\Rightarrow [SO_4^{2-}]=\dfrac{0,1}{1+\dfrac{h}{K_a}}$$ 
$$\Rightarrow h=\dfrac{10^{-14}}{h}+\dfrac{0,1h}{K_a+h}+\dfrac{0,2K_a}{K_a+h}$$ 
$$\Rightarrow h=0,1098$$ 
Vậy: 
$$[H^+]=0,1098M$$ 
$$[SO_4^{2-}]=\dfrac{0,1}{1+\dfrac{h}{K_a}}=0,01M$$ 
$$[HSO_4^-]=0,1-0,01=0,09M$$