Question

The concentration of Pb²⁺ in a solution saturated withPbBr₂(s) is 2.14 x 10^-2 M. Calculate K_sp for PbBr₂.

Short answer

Thevalue of K_sp can be found as $3.92\times {10}^{-5}$

Step-by-step solution

Given Information

The solubility can be given as$2.14\times {10}^{-2}mol/L$

Calculating the value of Ksp

The dissolution equilibrium can be denoted as:

$$\begin{gathered} PbB{r}_2\left(s\right)\iff P{b}^{+2}\left(aq\right)+2B{r}^{-}\left(aq\right) \\ {K}_{sp}=\left[P{b}^{+2}\right]{\left[B{r}^{-}\right]}^2 \\ \left[P{b}^{+2}\right]=smol/L \\ \left[B{r}^{-}\right]=2smol/L \\ {K}_{sp}=\left[s\right]{\left[2s\right]}^2=4s^3 \\ {K}_{sp}=4{\left(2.14\times {10}^{-2}\right)}^3 \\ {K}_{sp}=3.92\times {10}^{-5} \end{gathered}$$