Question

A solution contains 0.0150 M Pb2+(aq) and 0.0150 M Sr2+(aq) . If you add SO2−4(aq) , what will be the concentration of Pb2+(aq) when SrSO4(s) begins to precipitate?

Answer 1

Answer:

$\large \boxed{1.10 \times 10^{-3}\text{ mol/L}}$

Explanation:

1. Concentration of SO₄²⁻

SrSO₄(s) ⇌ Sr²⁺(aq) +SO₄²⁻(aq); Ksp = 3.44 × 10⁻⁷

                   0.0150          x

$K_{sp} =\text{[Sr ^{2+} ][SO _{4}^{2-} ]} = 0.0150x = 3.44 \times 10^{-7}\\x = \dfrac{3.44 \times 10^{-7}}{0.0150} = \mathbf{2.293 \times 10^{-5}} \textbf{ mol/L}$

2. Concentration of Pb²⁺

PbSO₄(s) ⇌ Pb²⁺(aq) + SO₄²⁻(aq); Ksp = 2.53 × 10⁻⁸

                        x          2.293 × 10⁻⁵

$K_{sp} =\text{[Pb ^{2+} ][SO _{4}^{2-} ]} = x \times 2.293 \times 10^{-5} = 2.53 \times 10^{-8}\\\\x = \dfrac{2.53 \times 10^{-8}}{2.293 \times 10^{-5}} = \mathbf{1.10 \times 10^{-3}} \textbf{ mol/L}\\\\\text{The concentration of Pb ^{2+} is \large \boxed{\mathbf{1.10 \times 10^{-3}}\textbf{ mol/L}} }$