Question

A$\mathbf{35}\mathbf{}\mathbf{\text{mL}}$solution of$\mathbf{0}\mathbf{.}\mathbf{075}\mathbf{}\mathbf{M}\mathbf{}{\mathbf{CaCl}}_{\mathbf{2}}$is mixed with$\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{mL}$of$\mathbf{0}\mathbf{.}\mathbf{090}\mathbf{}\mathbf{M}\mathbf{}{\mathbf{BaCl}}_{\mathbf{2}}$.

(a) If aqueous$\mathbf{KF}$is added, which fluoride precipitates first?

(b) Describe how the metal ions can be separated using$\mathbf{KF}$to form the fluorides.

(c) Calculate the fluoride ion concentration that will accomplish the separation.

Short answer

(a) If aqueous$\text{KF}$ is added, the ion which precipitates first is.

(b) The separation of metal ions using $\text{KF}$is done by $\text{KF}$adding slowly.

(c) The fluoride ion concentration slightly above $2.5\times {10}^{-5}\mathrm{M}$will accomplish the separation.

Step-by-step solution

Concept Introduction.

${\mathbf{Q}}_{\mathbf{sp}}$- ion-product expression;${\mathbf{Q}}_{\mathbf{sp}}$value is obtained when the concentrations of ions formed by dissolution of some compound are multiplied. When the solution is saturated${\mathbf{Q}}_{\mathbf{sp}}$value is calledrole="math" localid="1663283534071" ${\mathbf{K}}_{\mathbf{sp}}$value (solubility-product constant).

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

Solid and liquid state is not included in the${\mathbf{K}}_{\mathbf{sp}}$equations.

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

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

Information Provided.

• Volume of ${\text{CaCl}}_2$is: $35\mathrm{mL}=0.035\mathrm{L}$
• Moles of ${\text{CaCl}}_2$is: $0.075\text{M}$
• Volume of ${\text{BaCl}}_2$is: $25\mathrm{mL}=0.025\mathrm{L}$
• Moles of ${\text{BaCl}}_2$is: $0.090\text{M}$

Addition of KF

(a)

When is $\mathrm{KF},{\mathrm{CaF}}_2$and data-custom-editor="chemistry" ${\mathrm{BaF}}_2$are added following changes takes place –

$\begin{array}{c}{\mathrm{CaF}}_2\iff {\mathrm{Ca}}^{2+}+2{\mathrm{F}}^{-}\\ {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ca}}^{2+}\right]{\left[{\mathrm{F}}^{-}\right]}^2=3.2\times {10}^{-12}\\ {\mathrm{BaF}}_2\iff B{\mathrm{a}}^{2+}+2{\mathrm{F}}^{-}\\ {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ba}}^{2+}\right]{\left[{\mathrm{F}}^{-}\right]}^2=1.5\times {10}^{-6}\end{array}$

Now calculate fluoride concentrations –

$\begin{array}{c}{\mathrm{CaF}}_2:\left[{\mathrm{F}}^{-}\right]=\sqrt{\frac{3.2\times {10}^{-11}}{\left[{\mathrm{Ca}}^{2+}\right]}}\\ \left[{\mathrm{Ca}}^{2+}\right]=\frac{0.035\mathrm{L}\times 0.075\mathrm{M}}{0.035\mathrm{L}+0.025\mathrm{L}}=0.0525\mathrm{M}\\ \left[{\mathrm{F}}^{-}\right]=2.7\times {10}^{-5}\mathrm{M}\end{array}$

$\begin{array}{c}{\mathrm{BaF}}_2:\left[{\mathrm{F}}^{-}\right]=\sqrt{\frac{1.5\times {10}^{-6}}{\left[{\mathrm{Ba}}^{2+}\right]}}\\ \left[{\mathrm{Ba}}^{2+}\right]=\frac{0.025\mathrm{L}\times 0.09\mathrm{M}}{0.035\mathrm{L}+0.025\mathrm{L}}=0.045\mathrm{M}\\ \left[{\mathrm{F}}^{-}\right]=6.3\times {10}^{-3}\mathrm{M}\end{array}$

Therefore, it can be seen that ${\mathrm{CaF}}_2$needs much lower concentration of ${\text{F}}^{-}$ions to begin precipitation, thus, ${\mathrm{CaF}}_2$precipitates first.

Separation of ions using KF.

(b)

In order to separate this metal cations by using $\text{KF}$, slowly add $\text{KF}$, but keep the concentration of $\left[{\mathrm{F}}^{-}\right]$slightly above than $2.5\times {10}^{-5}\mathrm{M}$and lower than $6.3\times {10}^{-3}\mathrm{M}$. Thus,${\text{CaF}}_2$ will precipitate and ${\mathrm{Ba}}^{2+}$will remain in the solution.

Therefore, slowly add $\text{KF}$to carry out separation.

Concentration of F-.

(c)

According to answer (b), the ${\text{F}}^{-}$concentration would be just a bit above than $2.5\times {10}^{-5}\mathrm{M}$.

Therefore, the concentration should be above than $2.5\times {10}^{-5}\mathrm{M}$.