The first and second ionization constants of H 2 X are 2 . 5 × 10 - 8 and 1 . 0 × 10 - 13 respectively. The concentration of X 2 - in 0 . 1 ? M H 2 X solution is______ × 10 - 15 ? M . (Nearest Integer) [2026]

Question

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 $0.1 M H_{2} X$ solution is______ $\times 10^{- 15} M$ . (Nearest Integer) [2026]

Smart Achievers · Accepted Answer

(100)

$\begin{matrix} H_{2} X & ⇌ & H^{+} & + & {HX}^{-} \\ 0.1 - x & x + y & x - y \end{matrix}$

$2.5 \times 10^{- 8} = \frac{(x + y) (x - y)}{0.1 - x}$

$\begin{matrix} {HX}^{-} & ⇌ & H^{+} & + & X^{2 -} \\ x - y & x + y & y \end{matrix}$

$1 \times 10^{- 13} = \frac{(x + y) (y)}{x - y}$

$Approximate : K_{a 1} ≫ K_{a 2} \Rightarrow So x ≫ y$

$x + y \approx x, x - y \approx x$

$10^{- 13} = \frac{x \cdot y}{x}$

$y = 10^{- 13}$

$[X^{2 -}] = 10^{- 13}$

$[X^{2 -}] = 100 \times 10^{- 15}$

Solution

Q.
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 $0.1 M H_{2} X$ solution is______ $\times 10^{- 15} M$ . (Nearest Integer) [2026]

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Solution

Q. The first and second ionization constants of H2X are 2.5×10-8 and 1.0×10-13 respectively. The concentration of X2- in 0.1 M H2X solution is______×10-15 M. (Nearest Integer) [2026]

Ans. (100) H2X⇌H++HX-0.1-xx+yx-y 2.5×10-8=(x+y)(x-y)0.1-x HX-⇌H++X2-x-yx+yy 1×10-13=(x+y)(y)x-y Approximate : Ka1≫Ka2⇒So x≫y x+y≈x, x-y≈x 10-13=x·yx y=10-13 [X2-]=10-13 [X2-]=100×10-15

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Q.
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 $0.1 M H_{2} X$ solution is______ $\times 10^{- 15} M$ . (Nearest Integer) [2026]