Solveeit Logo
Login

Question

Question: The number of S<sup>2–</sup> ions present in 1L of 0.1(M) H<sub>2</sub>S solution having [H<sup>+</s...

The number of S2– ions present in 1L of 0.1(M) H2S solution having [H+] = 0.1 (M) is

Given H2S⇄\rightleftarrows2H+ + S2–Ka = 1.1 × 10–21)

A

6.625 × 103

B

6.625 × 104

C

6.625 × 105

D

6.625 × 106

Answer

6.625 × 103

Explanation

Solution

H2S ⇄\rightleftarrows2H+ + S2–

Ka = 1.1 × 10–21 = [H+]2[S2−][H2S]\frac{\lbrack H^{+}\rbrack^{2}\lbrack S^{2 -}\rbrack}{\lbrack H_{2}S\rbrack} = 10−2×[S2−]0.1\frac{10^{- 2} \times \lbrack S^{2 -}\rbrack}{0.1}

or [S2–] = 1.1 × 10–20 (m)

\ no. of S2– ions = 1.1 × 10–20 × 6.023 × 1023

= 1.1 × 6.023 × 103 = 6.625 × 103