Question

The minimum volume of water required to dissolve $0.1$g lead(II) chloride to get a saturated solution ($K_{sp}$ of $PbC{l}_2=3.2\times {10}^{-8}$; atomic mass of Pb$=207u$) is:

• A. $0.18$ L
• B. $17.98$ L
• C. $1.798$ L
• D. $0.36$ L

Answer 1

The correct option is A $0.18$ L

$Ksp=3.2\times {10}^{-8}$

$PbC{l}_2\rightleftharpoons P{b}^{2+}+2C{l}^{-}$

$K_{sp}=\left(S\right)(2S{)}^2$

$K_{sp}=4S^3$

$s^3=\frac{3.2\times {10}^{-8}}{4}$

$s^3=8\times {10}^{-9}$

$s=2\times {10}^{-3}$

$MolarmassofPbC{l}_2=278g/mol$

By using Molarity formula,

$\therefore 2\times {10}^{-3}=\frac{0.1}{278\times x}$

$\therefore x=\frac{0.1}{278\times 2\times {10}^{-3}}$

$=\frac{100}{278\times 2}$

$x=0.18L$