The first and second ionization constants of H_2X are 2.5 10^-8 and 1.0 10^-13 respectively. The concentration of X^2- in 0.1 MH_2X solution is \_ \_ \_ \_ 10^-15M. (Nearest Integer)

Question

Accepted Answer

Correct answer: 100. (100) For a weak diprotic acid H_2X, the ionization steps are: H_2X ⇌H^++HX^-with K_a1=2.5 10^-8 HX^- ⇌H^++X^2- with K_a2=1.0 10^-13 Since K_a1 ≫K_a2, the concentration of H^+is primarily determined by the first ionization. (H^+ ) ≈ K_a1 C= 2.5 10^-8 0.1= 2.5 10^-9=5 10^-5M Also, from the first ionization, (HX^- ) ≈ (H^+ )=5 10^-5M For the second ionization: K_a2= (H^+ ) (X^2- ) (HX^- ) Substituting the values: 1.0 10^-13= (5 10^-5) (X^2- )5 10^-5 Thus, (X^2- )=K_a2=1.0 10^-13M To express this in the form x 10^-15M : 1.0 10^-13=100 10^-15M The value is 100 .

$The first and second ionization constants of H_{2} X are 2.5 \times 10^{- 8} and 1.0 \times 10^{- 13} respectively. The concentration of X^{2 -} in 0.1 MH_{2} X solution is____\times 10^{- 15} M . (Nearest Integer)$

Solution

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