Tính nồng độ các ion H+ , HSO4- , SO42- trong dung dịch H2SO4 0,1M . Cho K2 = 1,2.10-2

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: 

$[H^+]=0,1098M$ 

$[SO_4^{2-}]=\dfrac{0,1}{1+\dfrac{h}{K_a}}=0,01M$ 

$[HSO_4^-]=0,1-0,01=0,09M$