Question

$\mathbf{\text{NaCl}}$is purified by adding$\mathbf{\text{HCl}}$to a saturated solution of$\mathbf{NaCl}\mathbf{(}\mathbf{317}\mathbf{g}\mathbf{/}\mathbf{L}\mathbf{)}$. When$\mathbf{28}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{mL}$of$\mathbf{8}\mathbf{.}\mathbf{65}\mathbf{}\mathbf{M}\mathbf{}\mathbf{HCl}$ is added to$\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{\text{L}}$ of saturated solution, what mass (g) of pure$\mathbf{NaCl}$ precipitates?

Short answer

No sodium chloride will precipitate because the reaction quotient is less than the solubility product constant.

Step-by-step solution

Definition of Concept.

Capacity: The amount of heat energy required to raise the temperature of a body by a certain amount is known as heat capacity. The amount of heat in joules required to raise the temperature 1 Kelvin is known as heat capacity (symbol: C) in SI units.

 Calculate the mass (g) of pure NaCI precipitates.

The moles of chloride ions and the solubility product constant of sodium chloride are calculated as follows:

$\begin{array}{c}\mathrm{Concentration}\left(\mathrm{M}\right)\mathrm{of}\\ \mathrm{NaCl}=\left(\frac{317\mathrm{gNaCl}}{\mathrm{L}}\right)\left(\frac{1\mathrm{molNaCl}}{58.{44}_{\mathrm{gNaCl}}}\right)\\ =5.42436687\mathrm{MNaCl}\\ {\mathrm{NaCl}}_{\left(\mathrm{s}\right)}\stackrel{}{\rightleftarrows }{\mathrm{Na}}_{\left(\mathrm{aq}\right)}^{+}{+\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}\\ {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Na}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]={\mathrm{S}}^2=(5.42436687{)}^2=29.42375594=29.4\end{array}$

Moles of ${\mathrm{Cl}}^{-}$initially

$=\left(\frac{5.42436687\mathrm{molNaCl}}{\mathrm{L}}\right)(0.100\mathrm{L})\left(\frac{1\mathrm{molCl}}{1\mathrm{molNaCl}}\right.)=0.542436687{\mathrm{molCl}}^{-}$

Moles of ${\text{Cl}}^{-}$is added $=\left(\frac{8.65\mathrm{mol}\mathrm{HCl}}{\mathrm{L}}\right)\left(\frac{{10}^{-3}{\mathrm{L}}^2}{1\mathrm{mL}}\right)(28.5\mathrm{mL})\left(\frac{\mathrm{I}\mathrm{mol}{\mathrm{Cl}}^{-}}{1\mathrm{mol}{\mathrm{HCl}}^2}\right.)=0.246525\mathrm{mol}{\mathrm{Cl}}^{-}$$0.100\text{L}$

of saturated solution contains $0.542\text{mol}$each ${\text{Na}}^{+}$and ${\text{Cl}}^{-}$to which one adding $0.466525\text{mol}$of additional chloride ions from hydrogen chloride.

Volume of mixed solutions$=0.100\mathrm{L}+(28.5\mathrm{mL})\left({10}^{-3}\mathrm{L}/1\mathrm{mL}\right)=0.1285\mathrm{L}$

Molarity of in mixture $=\left[(0.542436687+0.246525)\mathrm{mol}\mathrm{C}{\mathrm{l}}^{-}\right]/(0.1285\mathrm{L})$

$=6.13978\mathrm{M}{\mathrm{Cl}}^{-}$

Molarity of in mixture$=\left(0.542436687\mathrm{mol}{\mathrm{Na}}^{+}\right)/(0.1285\mathrm{L})$

$\begin{array}{c}=4.22130{\mathrm{MNa}}^{+}\\ {\mathrm{Q}}_{\mathrm{sp}}=\left[{\mathrm{Na}}^{+}\right]\left[{\mathrm{Cl}}^{-}\right]=(4.22130\left)\right(6.13978)=25.9179=25.9\end{array}$

Therefore, No sodium chloride will precipitate because the reaction quotient is less than the solubility product constant.