Q19.152 CP - Chemistry: Molecular Nature Of Matter And Change

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?

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.