Question

a. Calculate the molar solubility of ${\mathbf{SrF}}_{\mathbf{2}}$ in water,ignoring the basic properties of ${\mathbf{F}}^{\mathbf{-}}$. (For ${\mathbf{SrF}}_{\mathbf{2}}$,${\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{.}$)
b. Would the measured molar solubility of be greater than or less than the value calculated in part a? Explain.
c. Calculate the molar solubility of${\mathbf{SrF}}_{\mathbf{2}}$ in a solution buffered at $\mathbf{pH}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}$.(${\mathbf{K}}_{\mathbf{a}}$, for$\mathbf{HF}$ is $\mathbf{7}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$.)

Short answer

1. The molar solubility of${\mathbf{SrF}}_{\mathbf{2}}$ in water is.
2. The measured molar solubility of${\mathbf{SrF}}_{\mathbf{2}}$ begreater than the value calculated in part (a).
3. The molar solubility of in a solution buffered at is .

Step-by-step solution

Introduction to the Concept

The solubility product constant is a constant that determines the solubility equilibrium that occurs between a substance in its solid-state and its ions in the aqueous state. It has the following general expression:

${\mathbf{A}}_{\mathbf{x}}{\mathbf{B}}_{\mathbf{y}}\left(S\right)\mathbf{\rightleftharpoons }{\mathbf{xA}}^{\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{yB}}^{\mathbf{-}}\left(\mathrm{aq}\right)$

${\mathbf{K}}_{\mathbf{SP}}\mathbf{=}{\left[A^{+}\right]}^{\mathbf{x}}{\left[B^{-}\right]}^{\mathbf{Y}}$

The solubility product constant is,${\mathrm{K}}_{\mathrm{SP}}$ and the concentration is represented by square brackets.

Subpart (a) Determination of molar solubility of SrF2 in water

${\mathrm{SrF}}_2\left(\mathrm{s}\right)\to {\mathrm{Sr}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{F}}^{-}\left(\mathrm{aq}\right)$

Initial	-	-
Eqm	s	2s

$\begin{array}{c}{\text{K}}_{\text{SP}}\text{=}\left[{\text{Sr}}^{\text{2+}}\right]{\left[{\text{F}}^{\text{-}}\right]}^{\text{2}}\\ {\text{7.9×10}}^{\text{-10}}\text{=s×}{\left(\text{2s}\right)}^{\text{2}}\\ =4s^3\end{array}$

Therefore, the value of s is

$\begin{array}{l}s={\left(\frac{7.9\times {10}^{-10}}{4}\right)}^{\frac{1}{3}}\\ s=5.8\times {10}^{-4}\text{M}\end{array}$

Therefore, the molar solubility is $5.8\times {10}^{-4}\mathrm{M}$.

Subpart (b) Determination of molar solubility of SrF2.

${\mathrm{SrF}}_2\left(\mathrm{s}\right)\to {\mathrm{Sr}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{F}}^{-}\left(\mathrm{aq}\right)$

The F-ion is a weak acid's conjugate base. As a result, fluoride is a more powerful base than water. Fluoride interacts with water in the following way.

${\mathrm{F}}^{-}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to \text{HF}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$

More${\mathrm{SrF}}_2$ dissolves to replace fluoride ions when the concentration of F- falls.

Therefore, the molar solubility computed in part (a) is higher.

Subpart (c) Determination of the molar solubility of SrF2 in a solution buffered at pH=2.00.

Because F-depicts the basic qualities, certain F-are transformed to HF.

If $Sr{F}_2$has a molar solubility of $\mathrm{s}\mathrm{mol}/\mathrm{L}$.

So, the total F-concentration $=2s$.

$2\mathrm{s}=\left[{\mathrm{F}}^{-}\right]+\left[\mathrm{HF}\right]$

$\mathrm{HF}\to {\mathrm{H}}^{+}+{\mathrm{F}}^{-}$

As, $\left[{\mathrm{H}}^{+}\right]=1.0\times {10}^{-2}\text{M}$

$\begin{array}{c}{\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{F}}^{-}\right]}{\left[\mathrm{HF}\right]}\\ 7.2\times {10}^{-4}=\frac{{10}^{-2}\left[{\mathrm{F}}^{-}\right]}{\left[\mathrm{HF}\right]}\end{array}$

So,$\left[\mathrm{HF}\right]=14\left[{\mathrm{F}}^{-}\right]$

Putting the values,

$2\mathrm{s}=\left[{\mathrm{F}}^{-}\right]+\left[\mathrm{HF}\right]$

and, $\left[{\mathrm{F}}^{-}\right]=\frac{2\mathrm{s}}{15}$

Now,

$\begin{array}{c}{\mathrm{K}}_{\mathrm{SP}}=\left[{\mathrm{Sr}}^{2+}\right]{\left[{\mathrm{F}}^{-}\right]}^2\\ 7.9\times {10}^{-10}=\mathrm{s}\times {\left(\frac{2\mathrm{s}}{15}\right)}^2\\ =\frac{4{\mathrm{s}}^3}{225}\end{array}$

The molar solubility,

$\begin{array}{l}\mathrm{s}={\left(\frac{7.9\times {10}^{-10}\times 225}{4}\right)}^{\frac{1}{3}}\\ \mathrm{s}=3.5\times {10}^{-3}\text{M}\end{array}$

Therefore, the molar solubility of${\mathbf{SrF}}_{\mathbf{2}}$ in a solution is $3.5\times {10}^{-3}\mathrm{M}$.