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:
- A 100
- B 10
- C 1
- D 0.1
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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