JEE Main 2022 · 25 Jun · Shift 2 · Chemistry Q64

ChemistrySingle correct+4 / -1Question 64 of 90

The $K_{sp}$ for bismuth sulphide ( $\mathrm{Bi_2S_3}$ ) is $1.08\times10^{-73}$ . The solubility of $\mathrm{Bi_2S_3}$ in mol L $^{-1}$ at 298 K is

1.0\times10^{-15}

2.7\times10^{-12}

3.2\times10^{-10}

4.2\times10^{-8}

Answer and worked solution

Official answer: A

Solubility product. $\mathrm{Bi_2S_3\rightleftharpoons2Bi^{3+}+3S^{2-}}$ : $K_{sp}=(2s)^2(3s)^3=108s^5$ .

Solve. $s^5=10^{-75}$ , so $s=1.0\times10^{-15}$ mol L $^{-1}$ .

(A)

Topics:Solubility product

JEE Main 2022 · 25 Jun · Shift 2, Chemistry question 64