Q19.130CP - Chemistry: Molecular Nature Of Matter And Change

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$\mathbf{\text{1 L}}$  of aqueous solution.

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

 $\begin{array}{ll}\mathbf{\text{ (I) Pure water }}& \mathbf{\text{ (II) }}\mathbf{0}\mathbf{.01}\mathbf{M}\mathbf{\text{HF}}\end{array}$

(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?

Step-by-Step Solution

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 1: 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 ( .$\left(V\right)0.01\text{\hspace{0.17em}M\hspace{0.17em}\hspace{0.17em}Ca}{\left(\text{OH}\right)}_2$