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JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)

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 ______\( \times 10^{-15} M\). The value of Y is:

  1. A 100
  2. B 10
  3. C 1
  4. D 0.1
Verified Solution

Answer & Solution

Correct Answer

(A) 100

Step-by-step Solution

Detailed explanation

\(H _2 X \rightleftharpoons H ^{+}+ HX ^{-}\), \(0.1-x \quad x+y \quad x-y\) \(2.5 \times 10^{-8}=\frac{(x+y)(x-y)}{0.1-x}\) \(HX ^{-} \rightleftharpoons H ^{+}+ X ^{2-}\) \(x-y \quad x+y \quad y\) \(1 \times 10^{-13}=\frac{( x + y )( y )}{ x - y }\) Approximate :…
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