Question

A solution containing lead(ii) nitrate is mixed with one containing sodium bromide to form a solution that is 0.0140 m in pb(no3)2 and 0.0024 m in nabr. what is the value of q for the insoluble product? express the reaction quotient to three significant figures.

Answer 1

Answer:

$Q_{\text{sp}} = 8.06 \times 10^{-8}$

Explanation:

Since all sodium salts and all nitrates are soluble, the insoluble product must be lead(II) bromide.

The equation for the equilibrium is

                     PbBr₂ ⇌ Pb²⁺  +  Br⁻

I/mol·L⁻¹:                    0.0140  0.0024

$Q_{\text{sp}}$ = [Pb²⁺][Br⁻]²

[Pb²⁺] = 0.0140 mol·L⁻¹

[Br⁻] = 0.0024 mol·L⁻¹

$Q_{\text{sp}} = 0.0140 \times 0.0024^{2}$

$Q_{\text{sp}} = 8.06 \times 10^{-8}$