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

`\large \boxed{1.10 * 10^(-3)\text{ mol/L}}`

Step-by-step 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 * 10^(-7)\\x = (3.44 * 10^(-7))/(0.0150) = \mathbf{2.293 * 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 * 2.293 * 10^(-5) = 2.53 * 10^(-8)\\\\x = (2.53 * 10^(-8))/(2.293 * 10^(-5)) = \mathbf{1.10 * 10^(-3)} \textbf{ mol/L}\\\\\text{The concentration of Pb$^(2+)$ is $\large \boxed{\mathbf{1.10 * 10^(-3)}\textbf{ mol/L}}$}`