Chapter 17, Problem 58QP

A volume of 75 mL of 0.060 M NaF is mixed with 25 mL of $\text{0}\text{.15}M\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$. Calculate the concentrations in the final solution of ${\text{NO}}_3^{-},{\text{Na}}^{\text{+}},S{r}^{2+},{\text{ and F}}^{\text{-}}\left(K_{\text{sp}}{\text{ for SrF}}_{\text{2}}=2.0\times {10}^{-10}.\right)$

Interpretation Introduction

Interpretation:

Thefinal concentration of ${\text{NO}}_{\text{3}}{}^{-}$, ${\text{Na}}^{\text{+}}$, ${\text{Sr}}^{\text{2+}}$, and ${\text{F}}^{\text{-}}$ are to be calculated with given volume and concentration of $\text{NaF}$ and $\text{Sr}{\left({\text{NO}}_3\right)}_2$ and $K_{\text{sp}}$ value of ${\text{SrF}}_2$.

Concept introduction:

The amount of solute dissolved in a given volume of the solvent to form a saturated solution at a given temperature is termed as solubility of the solute in the solvent, at that temperature.

The solubility product of the sparingly soluble salt is given as the product of the concentration of the ions raised to the power equal to the number of times the ion occurs in the equation, after dissociation of the electrolyte.

Number of moles of solute dissolved per litre of solution is called molar solubility.

The number of moles of compound are calculated by the expression:

$\text{Number of moles}=\text{Concentration}\times \text{Volume of sol}$

The mole is converted into the concentration by dividing with volume.

$\text{Concentration}=\frac{\text{Moles}}{\text{Volume}}$

At a given temperature, the product of molar concentrations of the ions of a salt present in the solution is known as the solubility product of the salt. It is represented by $K_{sp}$.

Higher is the value of solubility product of a salt, higher is its solubility.

The presence of common ions in the solution decreases the solubility of a given compound.

For a general reaction: $\text{AB}\left(s\right)\rightleftharpoons {\text{A}}^{+}\left(aq\right)+{\text{B}}^{-}\left(aq\right)$

The solubility product can be calculated by the expression as: $K_{\text{sp}}{\text{= [A}}^{\text{+}}{\text{][B}}^{-}\text{]}$

Here, $K_{\text{sp}}$ is the solubility product constant and $sp$ stands for solubility product.

The molar solubility of a compound is directly proportional to the number of molecules present in the given amount of solvent.

Answer to Problem 58QP

Solution: The final concentration of $\left[{\text{NO}}_3^{-}\right]\text{ is }0.076\text{M}$, of $\left[{\text{Na}}^{+}\right]\text{ is }0.045\text{M}$, of $\left[{\text{Sr}}^{2+}\right]\text{ is }0.016\text{M}$ and of $\left[{\text{F}}^{-}\right]\text{ is }1.1\times {10}^{-4}\text{M}$.

Explanation of Solution

Given information: The solubility product of ${\text{SrF}}_2$ is $2\times {10}^{-10}$, the concentration of $\text{NaF}$ is $0.06\text{M }$, volume of $\text{NaF}$ is $75\text{mL}$, the concentration of $\text{Sr}{\left({\text{NO}}_3\right)}_2$ is $0.15\text{M }$, and volume of $\text{Sr}{\left({\text{NO}}_3\right)}_2$ is $25\text{mL}$.

The equation of ionic reactionis as follows:

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

Moles of ${\text{Sr}}^{2+}$ ion are calculated using the following expression:

${\text{Moles of Sr}}^{2+}={\text{Concentration of Sr}}^{2+}\times \text{Volume of sol}$

Substitute the value of concentration and volume in the above expression.

$\begin{array}{c}{\text{Moles of Sr}}^{2+}=\left(25\text{mL}\times \frac{0.15\text{mol}}{1000\text{mL sol}}\right)\\ =0.0038\text{mole}\end{array}$

Moles of ${\text{F}}^{-}$ ion are calculated using the following expression:

${\text{Moles of F}}^{-}={\text{Concentration of F}}^{-}\times \text{Volume of sol}$

Substitute the value of concentration and volume in the above expression,

$\begin{array}{l}{\text{Moles of F}}^{-}=\left(75\text{mL}\times \frac{0.060\text{mol}}{1000\text{mL sol}}\right)\\ =0.0045\text{mole}\end{array}$

Using the stoichiometry of the balanced equation, moles of ${\text{F}}^{\text{-}}$ required to react with ${\text{Sr}}^{\text{2+}}$ are twice that of Sr²⁺. Divide the moles of reactant with its stoichiometry coefficient and the one with the lowest ratio, is the limiting reagent.

The limiting reagent of the reactionis calculated as follows:

The stoichiometry coefficient of ${\text{Sr}}^{\text{2+}}$ is 1.

The ratio is as follows:

$\frac{\text{Mole}}{\text{Stoichemetric coefficient}}=\frac{0.0038}{1}$

The stoichiometry coefficient of ${\text{F}}^{-}$ is 2.

The ratio is as follows:

$\begin{array}{l}\frac{\text{Mole}}{\text{Stoichemetric coefficient}}=\frac{0.0045}{2}\\ =0.00225\end{array}$

Compare both ratios. The one with the lowest ratio is the limiting reagent.

Thus, ${\text{F}}^{-}$ is the limiting reagent.

Summarize the moles at the equilibrium as follows:

$\begin{array}{l}{\text{Sr}}^{2+}\left(aq\right)+{\text{ 2F}}^{-}\left(aq\right)\to {\text{ SrF}}_2\left(s\right)\\ \begin{array}{cccc}\text{Initial}\left(\text{mol}\right)& 0.0038& 0.0045& 0\\ \text{Change}\left(\text{mol}\right)& -0.00225& 0.0045& +0.00225\\ \text{Equilibrium}\left(\text{mol}\right)& 0.00155& 0& 0.00225\end{array}\end{array}$

Total volume of the solution is $100\text{mL}\text{=}\text{0}\text{.1}\text{L}$

The mole is converted into the concentration by dividing with volume.

$\text{Concentration}=\frac{\text{Moles}}{\text{Volume}}$

Summarise the concentration at the equilibrium as follows:

Consider s be the molar solubility.

$\begin{array}{l}{\text{SrF}}_2\left(s\right)\rightleftharpoons {\text{Sr}}^{2+}\left(aq\right)+2{\text{F}}^{-}\left(aq\right)\\ \begin{array}{cccc}\text{Initial}\left(M\right)& 0.0225& 0.0155& 0\\ \text{Change}\left(M\right)& -s& +s& +2s\\ \text{Equilibrium}\left(M\right)& 0.0225-s& 0.0155+s& 2s\end{array}\end{array}$

The equilibrium expression for a reaction is written as:

${\text{K}}_{\text{sp}}{\text{= [Sr}}^{\text{2+}}{\text{][F}}^{\text{-}}{\text{]}}^{\text{2}}$

Here, ${\text{K}}_{\text{sp}}$ is solubility product constant, ${\text{Sr}}^{2+}$ is the concentration of strontium ion and ${\text{F}}^{-}$ is the concentration fluoride ion.

Substitute the value of $K_{\text{sp}}$, ${\text{Ag}}^{+}$, and ${\text{Cl}}^{-}$ in the above expression,

$K_{\text{sp}}=\left(0.0115+s\right){\left(2s\right)}^2$

The value s is very small as compared to 0.0115. It can be neglected.

$\begin{array}{c}{K}_{\text{sp}}=\left(0.0115\right)4s^2\\ 2\times {10}^{-10}=\left(0.0115\right)4s^2\\ s={\left(\frac{2\times {10}^{-10}}{0.046}\right)}^{\frac{1}{2}}\\ s={\left(43.47\times {10}^{-10}\right)}^{\frac{1}{2}}\\ =6.6\times {10}^{-5}\text{M}\end{array}$

Concentration of $\left[{\text{Sr}}^{2+}\right]$ is calculated as follows:

$\begin{array}{l}\left[{\text{Sr}}^{2+}\right]=0.0155+s\\ =0.016\text{M}\end{array}$

Concentration of $\left[{\text{F}}^{-}\right]$ is calculated as follows:

$\begin{array}{l}\left[{\text{F}}^{-}\right]=2s\\ =1.32\times {10}^{-4}\text{M}\end{array}$

Both sodium and nitrate ions are the spectator ions because it does not affect the equilibrium. It exits in same form in both side of the reaction.

Concentration of $\left[{\text{NO}}_3^{-}\right]$ is calculated as follows:

$\begin{array}{l}\left[{\text{NO}}_3^{-}\right]=\frac{2\left(0.0038\right)\text{mol}}{\left(0.10\text{L}\right)}\\ =0.076\text{M}\end{array}$

The concentration of $\left[{\text{Na}}^{+}\right]$ is calculated as follows:

$\begin{array}{l}\left[{\text{Na}}^{+}\right]=\frac{\left(0.0045\text{mol}\right)}{\left(0.1\text{L}\right)}\\ =0.045\text{ M}\end{array}$

Conclusion

The final concentration of $\left[{\text{NO}}_3^{-}\right]\text{ is }0.076\text{M}$, final concentration of $\left[{\text{Na}}^{+}\right]\text{ is }0.045\text{M}$, final concentration of $\left[{\text{Sr}}^{2+}\right]\text{ is }0.016\text{M}$, and final concentration of $\left[{\text{F}}^{-}\right]\text{ is }1.1\times {10}^{-4}\text{M}$ $\left[{\text{F}}^{-}\right]\text{ is }1.32\times {10}^{-4}\text{M}$.