Q19.147 CP - Chemistry: Molecular Nature Of Matter And Change

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.

Step-by-Step Solution

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 1: 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 called  ${\mathbf{K}}_{\mathbf{sp}}$ 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.

Step 2: 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: $0.090 \text{M}$ 

Step 3: 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.

Step 4: 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.

Step 5: 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 $2.5\times {10}^{-5} \mathrm{M}$.