Q19.76P - Chemistry: Molecular Nature Of Matter And Change

Question

Calculate the molar solubility of${\mathbf{\text{Ca(IO}}}_{\mathbf{\text{3}}}{\mathbf{\text{)}}}_{\mathbf{\text{2}}}$in 

(a) ${\mathbf{\text{0.060 M Ca(NO}}}_{\mathbf{\text{3}}}{\mathbf{\text{)}}}_{\mathbf{\text{2}}}$ and 

(b) ${\mathbf{\text{0.060 M NaIO}}}_{\mathbf{\text{3}}}$. (See Appendix C.)

Step-by-Step Solution

Answer

(a) The molar solubility of ${\text{Ca(IO}}_{\text{3}}{\text{)}}_{\text{2}}$in ${\text{0.060 M Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $\mathrm{S}=1.7\times {10}^{-3}\mathrm{M}$.

(b) The molar solubility of ${\text{Ca(IO}}_{\text{3}}{\text{)}}_{\text{2}}$ in ${\text{0.060 M NaIO}}_{\text{3}}$ is $\mathrm{S}=1.97\times {10}^{-4}\mathrm{M}$.

Concept Introduction

The solubility product is directly related to the maximum number of moles of the solute that can be dissolved in a litre of solution before the solution becomes saturated. Any additional solute precipitates out of a solution once it is saturated.

Q_sp - ion-product expression; Q_sp value is obtained when the concentrations of ions formed by dissolution of some compound are multiplied. When the solution is saturated Q_sp  value is called K_sp value (solubility-product constant).

${\mathbf{MX}}_{\mathbf{2}}\mathbf{\to }{\mathbf{M}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{X}}^{\mathbf{-}}$

The solid and liquid state is not included in the K_sp equations.

${\mathbf{\text{K}}}_{\mathbf{\text{sp}}}{\mathbf{\text{ = [M}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{][X}}}^{\mathbf{\text{ - }}}{\mathbf{\text{]}}}^{\mathbf{\text{2}}}$

S (Molar solubility) is equal to the concentration of one mol of the ion formed by the dissolution of some compound.

Solubility Ca(IO 3 ) 2 of in Ca(NO 3 ) 2

(a)

Write the equation of dissolution for ${\text{Ca(IO}}_{\text{3}}{\text{)}}_{\text{2}}$–

${\text{Ca(IO}}_{\text{3}}{\text{)}}_{\text{2}}\text{(s)}\iff {\text{Ca}}^{\text{2 + }}{\text{(aq) + 2IO}}_3^{\text{ - }}\text{(aq)}$

In this case, the initial concentration of ${\text{Ca}}^{\text{2 + }}$ion is $\text{0.06 M + S}$, as there is${\text{0.06 M Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ so the equilibrium concentrations will be –

$$\begin{gathered} {\text{[Ca}}^{\text{2 + }}\text{]=0.06M + S} \\ {\text{[IO}}_3^{\text{ - }}\text{]=2S} \end{gathered}$$

Rearranging the equation of ${\text{K}}_{\text{sp}}$and solving –

Square ${\text{[IO}}_3^{\text{ - }}\text{]}$ as there are $2$ moles of ${\text{IO}}_3^{\text{ - }}$in the equation.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]\left[{\mathrm{IO}}_3^{-}\right] \\ {\mathrm{K}}_{\mathrm{sp}}=(0.06\mathrm{M}+\mathrm{S})·(2\mathrm{S}{)}^2 \\ {\mathrm{K}}_{\mathrm{sp}}=7.1\times {10}^{-7} \\ \mathrm{S}=1.7\times {10}^{-3}\mathrm{M} \end{gathered}$$ 

Therefore, the value for solubility is obtained as $\mathrm{S}=1.7\times {10}^{-3}\mathrm{M}$.

Solubility of Ca(IO 3 ) 2 in NaIO 3

(b)

In this case, the initial concentration of ${\text{IO}}_3^{\text{2 -}}$ion is $\text{0.06 M + 2S}$, as there is ${\text{0.06 M NaIO}}_{\text{3}}$so the equilibrium concentrations will be –

$$\begin{gathered} \left[{\mathrm{Ca}}^{2+}\right]=\mathrm{S} \\ \left[{\mathrm{IO}}_3^{2-}\right]=0.06\mathrm{M}+2\mathrm{S} \end{gathered}$$

Rearranging the equation of K_sp and solving –

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]\left[{\mathrm{IO}}_3^{2-}\right] \\ {\mathrm{K}}_{\mathrm{sp}}=\mathrm{S}·(0.06\mathrm{M}+2\mathrm{S}{)}^2 \\ {\mathrm{K}}_{\mathrm{sp}}=7.1\times {10}^{-7} \\ \mathrm{S}=1.97\times {10}^{-4}\mathrm{M} \end{gathered}$$ 

Therefore, the value for solubility is obtained as $\mathrm{S}=1.97\times {10}^{-4}\mathrm{M}$.