Question

Calculate the concentration of$P{b}^{2+}$in each of thefollowing.

a. a saturated solution of$Pb{\left(OH\right)}_2$,$K_{sp}=1.2\times {10}^{-15}a$

b. a saturated solution of$Pb{\left(OH\right)}_2$buffered at$pH=13.00$

c. Ethylenediaminetetraacetate$\left(EDT{A}^{4-}\right)$is used as acomplexing agent in chemical analysis and has thefollowing structure:

Solutions of$EDT{A}^{4-}$are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The reaction of$EDT{A}^{4-}$with$P{b}^{2+}$is

$P{b}^{2+}\left(aq\right)+EDT{A}^{4-}\left(aq\right)\rightleftharpoons PbEDT{A}^{2-}\left(q\right)K=1.1\times {10}^{18}$
Consider a solution with 0.010 mole of$Pb{\left(N{O}_3\right)}_2$addedto 1.0 L of an aqueous solution buffered at$pH=13.00$and containing 0.050 M.$N{a}_{4}EDTA$ Does$Pb{\left(OH\right)}_2$precipitate from this solution?

Short answer

a.$s=\left[P{b}^{2+}\right]=6.7\times {10}^{-6}M$

b.$s=\left[P{b}^{2+}\right]=1.2\times {10}^{-13}M$

c. The precipitation does not occur.

Step-by-step solution

The concentration of a saturated solution of Pb(OH)2  when .Ksp=1.2×10−15

The equation can be denoted as,

$Pb{\left(OH\right)}_2\left(s\right)\rightleftharpoons P{b}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)$

As,

data-custom-editor="chemistry" $\begin{array}{c}{K}_{sp}=\left[P{b}^{2+}\right]{\left[O{H}^{-}\right]}^2\\ 1.2\times {10}^{-15}=s\times {\left(2\mathrm{s}\right)}^2=4s^3\end{array}$

And,

data-custom-editor="chemistry" $s=\left[P{b}^{2+}\right]=6.7\times {10}^{-6}M$

The concentration of a saturated solution of Pb(OH)2 buffered at.data-custom-editor="chemistry" pH = 13.00

As,

$\begin{array}{l}pH+pOH=14\\ 13+pOH=14\end{array}$

And,

$\begin{array}{l}pOH=-\log \left[O{H}^{-}\right]aspOH=1\\ 1=-\log \left[O{H}^{-}\right]\end{array}$

So,

$\left[O{H}^{-}\right]=0.1M$

Also,

$\begin{array}{c}{K}_{sp}=\left[P{b}^{2+}\right]\left[O{H}^{-}\right]\\ 1.2\times {10}^{-15}=s\times {\left(0.1\right)}^2\end{array}$

Hence,

$s=\left[P{b}^{2+}\right]=1.2\times {10}^{-13}M$

The precipitation for Pb(OH)2when Ethylenediaminetetraacetate is used as a complexing agent.

It can be denoted as,

$\begin{array}{c}K=\frac{\left[PbEDT{A}^{2-}\right]}{\left[P{b}^{2+}\right]\left[EDT{A}^{4-}\right]}\\ 1.1\times {10}^8=\frac{0.010-x}{x(0.040+x)}\end{array}$

And,

$x=\left[P{b}^{2+}\right]=2.3\times {10}^{-19}Ma$

The reaction quotient can be calculated as,

$\begin{array}{c}Q=\left[P{b}^{2+}\right]{\left[O{H}^{-}\right]}^2\\ =2.3\times {10}^{-19}\times {\left(0.1\right)}^2\\ Q=2.3\times {10}^{-21}\end{array}$

Hence,

$Q<K_{sp}$

Thus,the precipitation does not occur here.