Question

student wants to dissolve the maximum amount of${\mathbf{\text{CaF}}}_{\mathbf{2}}$ $\mathbf{\left(}{\mathbf{\text{K}}}_{\mathbf{\text{sp}}}{\mathbf{\text{=3.2×10}}}^{\mathbf{\text{-11}}}\mathbf{\right)}$to make$\mathbf{\text{1L}}$ of aqueous solution.

(a) Into which of the following solvents should she dissolve the salt?

localid="1664178101553" $\begin{array}{ll}\mathbf{\text{(I)Purewater}}& \mathbf{\text{(II)}}\mathbf{0}\mathbf{.01}\mathbf{M}\mathbf{\text{HF}}\end{array}$

(III)$\mathbf{0}\mathbf{.01}\mathbf{\text{\hspace{0.17em}M\hspace{0.17em}\hspace{0.17em}NaOH}}$ (IV)$\mathbf{0}\mathbf{.01}\mathbf{\text{\hspace{0.17em}M\hspace{0.17em}\hspace{0.17em}HCl}}$ (V)$\mathbf{0}\mathbf{.01}\mathit{M}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}Ca}}{\mathbf{\left(}\mathbf{\text{OH}}\mathbf{\right)}}_{\mathbf{2}}$

(b) Which would dissolve the least amount of salt?

Short answer

a) The required solution for more dissolving is(IV) .$0.01\text{\hspace{0.17em}M\hspace{0.17em}\hspace{0.17em}HCl}$

b) The requires solution for least dissolving is (V) .$\left(\text{V}\right)0.01\text{\hspace{0.17em}M\hspace{0.17em}\hspace{0.17em}Ca}{\left(\text{OH}\right)}_2$

Step-by-step solution

Definition of pH

pH is a measure of hydrogen ion concentration, which indicates whether a solution is acidic or alkaline.

Step 2: Which of the following solvents should dissolve the salt

a)

The common ion effect is defined as a shift in an ionic or solubility equilibrium away from production of an ion caused by the addition of that ion.

Take a look at the following for each watery solution:

(I) Water without impurities The clean water does not change since there is no common ion.

(II) The HF solution is lonized for $0.01\text{M\hspace{0.17em}HF}$to yield hydrogen and fluoride ions. In this instance, the F ion is a common ion. Due to the common ion effect $Ca{F}_2$dissolves substantially more in solution than pure water.

(III)$0.01\text{\hspace{0.17em}M\hspace{0.17em}\hspace{0.17em}NaOH}$the base equilibrium reaction shifts to the left when more hydroxide ions are added, resulting in more $\mathrm{F}$ions.

(IV) . $0.01\text{M\hspace{0.17em}\hspace{0.17em}HCl}$The ions in the HCI solution eliminate the OH ions. The base equilibrium process consumes F ions as a result, and moves to the right. This means that more $\text{CaF}2$dissolves in this acid solution than in pure water.

(V) $0.01\text{\hspace{0.17em}M\hspace{0.17em}\hspace{0.17em}Ca}\left(\text{OH}\right)2$. The $\text{Ca}{\left(\text{OH}\right)}_2$solution is lonized to yield calcium and hydroxide ions. In this circumstance, the ion $Ca2$ is a common ion. Due to the common ion effect, ${\text{CaF}}_2$is somewhat more soluble in solution than pure water. Adding more hydroxide ions causes the base equilibrium reaction to shift to the left, resulting in more F ions. As a result of the additional F ions, the equilibrium will shift to the left. As a result, CaF is somewhat dissolved.

Adding an acid enhances ${\text{CaF}}_2$solubility because weak acid anions react with H to shift the equilibrium to the right side. This shows that in the acidic solution,${\text{CaF}}_2$dissolves. To maximise solubility, a $0.01\text{\hspace{0.17em}M\hspace{0.17em}\hspace{0.17em}HCl}$solution should be employed. As a result, the solution required for additional dissolving is (IV) .$0.01\text{\hspace{0.17em}M\hspace{0.17em}\hspace{0.17em}HCl}$

Step 3: Which would dissolve the least amount of salt

b)

Calcium fluoride$\left({\text{CaF}}_2\right)$will be the least soluble in a$Ca{\left(OH\right)}_2$solution. The equilibrium will gradually shift to the left due to the common-ion effect.

As a consequence, the solution with the least amount of dissolving is ( .$\left(V\right)0.01\text{\hspace{0.17em}M\hspace{0.17em}\hspace{0.17em}Ca}{\left(\text{OH}\right)}_2$