Question

Question:Suppose 50.0 mL of a 0.0500 M solution of is mixed with 40.0 mL of a 0.200 M solution of ${\mathbf{NaIO}}_{\mathbf{3}}$at 25⁰C . Calculate the $\left[{\mathbf{Pb}}^{\mathbf{2}+}\right]$ and $\left[\mathbf{I}{{\mathbf{O}}_{\mathbf{3}}}^{-}\right]$ when the mixture comes to equilibrium. At this temperature, Ksp for$\mathbf{Pb}{\left({\mathbf{IO}}_{\mathbf{3}}\right)}_{\mathbf{2}}$is $\mathbf{2}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{13}}\mathbf{.}$

Short answer

Answer

Thus, at equilibrium,

$$\begin{gathered} \left[P{b}^{2+}\right]=7.15\times {10}^{-11} M \\ \left[I{O_3}^{-}\right]=0.0603 M \end{gathered}$$

Step-by-step solution

The relation between reaction quotient and solubility product  

During certain chemical double displacement reactions, sometimes a solid residue will be formed and it is called precipitate.

When

$$\begin{gathered} {\mathbf{Q}}_{\mathbf{sp}}\mathbf{<}{\mathbf{K}}_{\mathbf{sp}}\mathbf{,} \mathbf{precipitate} \mathbf{is} \mathbf{not} \mathbf{formed}\mathbf{.} \\ {\mathbf{Q}}_{\mathbf{sp}}\mathbf{=}{\mathbf{K}}_{\mathbf{sp}}\mathbf{,} \mathbf{equilibrium} \mathbf{position} \\ {\mathbf{Q}}_{\mathbf{sp}}\mathbf{>}{\mathbf{K}}_{\mathbf{sp}}\mathbf{,} \mathbf{precipitate} \mathbf{is} \mathbf{formed}\mathbf{.} \end{gathered}$$

Where,${\mathbf{Q}}_{\mathbf{sp}}$ is the reaction quotient, ${\mathbf{K}}_{\mathbf{sp}}\mathbf{,}$is the solubility product.

The value of reaction quotient

The concentration of lead (II) ion in the solution:

$$\begin{gathered} \left[P{b}^{2+}\right]=\frac{0.0500 M\times 0.050 L}{0.090 L} \\ =0.0277 M \end{gathered}$$

The concentration of iodate ion in the solution:

$$\begin{gathered} \left[I{O_3}^{-}\right]=\frac{0.200 M\times 0.040 L}{0.090 L} \\ =0.088 M \end{gathered}$$

The value of reaction quotient:

$$\begin{gathered} Q_{sp}=\left[P{b}^{2+}\right]{\left[I{O_3}^{-}\right]}^2 \\ =\left(0.0277\right){\left(0.088\right)}^2 \\ =2.14\times {10}^{-4} \end{gathered}$$

The value of :$K_{sp}=2.6\times {10}^{-13}$

Therefore,$Q_{sp}>K_{sp}$ .

Precipitation occurs.

Equilibrium concentrations

Construct ICE table:

Thus, at equilibrium,

$$\begin{gathered} \left[P{b}^{2+}\right]=7.15\times {10}^{-11} M \\ \left[I{O_3}^{-}\right]=0.0603 M \end{gathered}$$