Question

For which of the following mixtures will Ag2SO4(s) precipitate?

• 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.20 M AgNO3(aq)
• 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.30 M AgNO3(aq)
• 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.40 M AgNO3(aq)
• 150.0 mL of 0.10 M Na2SO4(aq) and 5.0 mL of 0.50 M AgNO3(aq)

Answer 1

Concepts and reason

The solubility product for the reaction is equilibrium constant, where the solid ionic compound dissociates into its ions in a solution. The solubility product is denoted as \(\mathrm{K}_{\mathrm{sp}}\). The solubility product value relates to the saturated solution and indicates the precipitate level of the compound. The formation precipitation starts when the ionic product exceeds the solubility product.

Fundamentals

The solubility product value of the compound depends on the concentrations of its ions in a solution. Example: \(\mathrm{AB}\) is a solid ionic compound. Precipitation: If the solubility product value is lesser than the concentration of the ions present in the solution, the compound precipitates in the solution.

 

The given solid ionic compound is \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) The equilibrium equation for the ionic compound is given below:

Therefore, the solubility product of the ionic compound is

$$ \begin{aligned} \mathrm{K}_{\mathrm{sp}} &=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{SO}_{4}^{2-}\right] \\ &=1.2 \times 10^{-5} \end{aligned} $$

The equilibrium equation has been written for the given solid ionic compound \(\left(\mathrm{Ag}_{2} \mathrm{SO}_{4}\right)\), and the solubility product has been derived from the equilibrium equation.

 

The balanced equation for the reaction of \(\mathrm{AgNO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) :

The initial concentration of \(\left[\mathrm{Ag}^{+}\right]\) ions in \(\mathrm{AgNO}_{3}\) is given below:

$$ \begin{array}{c} {\left[\mathrm{Ag}^{+}\right]_{\text {initial }}=\frac{5.0 \mathrm{~mL} \times \frac{0.20 \mathrm{mmol} \mathrm{AgNO}_{3}}{\mathrm{~mL}} \times \frac{1 \mathrm{molAg}^{+}}{150.0+5.0) \mathrm{mL}}}{1 \mathrm{molAgNO}_{3}}} \\ =0.00645 \mathrm{M} \end{array} $$

The initial concentration of \(\left[\mathrm{SO}_{4}^{2-}\right]\) ions in \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is given below:

$$ \begin{array}{c} {\left[\mathrm{SO}_{4}^{2-}\right]_{\text {initial }}=\frac{150.0 \mathrm{~mL} \times \frac{0.10 \mathrm{mmolNa}_{2} \mathrm{SO}_{4}}{\mathrm{~mL}} \times \frac{1 \mathrm{~mol} S O_{4}^{2-}}{1 \mathrm{molNa}_{2} \mathrm{SO}_{4}}}{(150.0+5.0) \mathrm{mL}}} \\ =0.0968 \mathrm{M} \end{array} $$

The ionic product of \(\mathrm{Ag}_{2} \mathrm{SO}_{3}\) is given below:

$$ \begin{aligned} \mathrm{K}_{\mathrm{IP}}=&\left[\mathrm{Ag}^{+}\right]_{\text {initial }}^{2}\left[\mathrm{SO}_{4}^{2-}\right]_{\text {initial }} \\ =&(0.00645)^{2}(0.0968) \\ &=4.0 \times 10^{-6} \end{aligned} $$

The ionic product \(\left(\mathrm{K}_{\mathrm{IP}}\right)\) is less than the solubility product \(\left(\mathrm{K}_{\mathrm{sp}}\right)\), \(\mathrm{K}_{\mathrm{IP}}<\mathrm{K}_{\mathrm{sp}} \Rightarrow\) precipitation does not takes place.

For the mixture, \(150.0 \mathrm{mLof} 0.10 \mathrm{MNa}_{2} \mathrm{SO}_{4}\) (aq) and 5.0mL of 0.20M AgNO \(_{3}\) (aq) precipitation does not take place.

The balanced equation for the reaction of \(\mathrm{AgNO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is given in which the two moles of \(\mathrm{AgNO}_{3}\) react with one mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) to produce the \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) precipitate. The concentration of silver ions \(\left[\mathrm{Ag}^{+}\right]\) and sulphate ions \(\left[\mathrm{SO}_{4}^{2-}\right]\) has been calculated to find the ionic product. The value of ionic product is less than that of the solubility product for which the precipitation does not takes place.

 

The balanced equation for the reaction of \(\mathrm{AgNO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) :

The initial concentration of \(\left[\mathrm{Ag}^{+}\right]\) ions in \(\mathrm{AgNO}_{3}\) is given below:

$$ \begin{array}{c} {\left[\mathrm{Ag}^{+}\right]_{\text {initial }}=\frac{5.0 \mathrm{~mL} \times \frac{0.30 \mathrm{mmol} \mathrm{AgNO}_{3}}{\mathrm{~mL}} \times 1 \mathrm{molAg}^{+}}{(150.0+5.0) \mathrm{mL}} \frac{1 \mathrm{~mol} \mathrm{AgrO}_{3}}{3}} \\ =0.00968 \mathrm{M} \end{array} $$

The initial concentration of \(\left[\mathrm{SO}_{4}^{2-}\right]\) ions in \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is given below:

$$ \begin{array}{c} {\left[\mathrm{SO}_{4}^{2-}\right]_{\text {initial }}=\frac{150.0 \mathrm{~mL} \times \frac{0.10 \mathrm{mmolNa}_{2} \mathrm{SO}_{4}}{\mathrm{~mL}} \times 1 \mathrm{~mol} S O_{4}^{2-}}{(150.0+5.0) \mathrm{mL}} \frac{1 \mathrm{molNa}_{2} \mathrm{SO}_{4}}{4}} \\ =0.0968 \mathrm{M} \end{array} $$

The ionic product of \(\mathrm{Ag}_{2} \mathrm{SO}_{3}\) is given below:

$$ \begin{aligned} \mathrm{K}_{\mathrm{IP}}=&\left[\mathrm{Ag}^{+}\right]_{\text {initial }}^{2}\left[\mathrm{SO}_{4}^{2-}\right]_{\text {initial }} \\ =&(0.00968)^{2}(0.0968) \\ &=9.1 \times 10^{-6} \end{aligned} $$

The ionic product \(\left(\mathrm{K}_{\mathrm{IP}}\right)\) is less than the solubility product \(\left(\mathrm{K}_{\mathrm{sp}}\right)\), \(\mathrm{K}_{\mathrm{IP}}<\mathrm{K}_{\mathrm{sp}} \Rightarrow\) precipitationdoesnottakesplace

For the mixture, \(150.0 \mathrm{mLof} 0.10 \mathrm{MNa}_{2} \mathrm{SO}_{4}\) (aq) and 5.0mL of 0.30M AgNO \(_{3}\) (aq) precipitation does not take place.

The concentration of silver ions \(\left[\mathrm{Ag}^{+}\right]\) and sulphate ions \(\left[\mathrm{SO}_{4}^{2-}\right]\) has been calculated to find the ionic product. The ionic product's value is less than that of the solubility product for which precipitation does not occur.

 

The balanced equation for the reaction of \(\mathrm{AgNO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) :

The initial concentration of \(\left[\mathrm{Ag}^{+}\right]\) ions in \(\mathrm{AgNO}_{3}\) is given below:

$$ \begin{array}{c} {\left[\mathrm{Ag}^{+}\right]_{\text {initial }}=\frac{5.0 \mathrm{~mL} \times \frac{0.40 \mathrm{mmolAgNO}_{3}}{\mathrm{~mL}} \times \frac{1 \mathrm{~mol} \mathrm{Ag}^{+}}{150.0+5.0) \mathrm{mL}} \mathrm{molAgNO}_{3}}{(150.0+5.0) \mathrm{mL}}} \\ =0.0129 \mathrm{M} \end{array} $$

The initial concentration of \(\left[\mathrm{SO}_{4}^{2-}\right]\) ions in \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is given below:

$$ \begin{array}{c} {\left[\mathrm{SO}_{4}^{2-}\right]_{\text {initial }}=\frac{150.0 \mathrm{~mL} \times \frac{0.10 \mathrm{mmolNa}_{2} \mathrm{SO}_{4}}{\mathrm{~mL}} \times 1 \mathrm{~mol} \mathrm{SO}_{4}^{2-}}{(150.0+5.0) \mathrm{mL}} \frac{1 \mathrm{molNa}_{2} \mathrm{SO}_{4}}{4}} \\ =0.0968 \mathrm{M} \end{array} $$

The ionic product of \(\mathrm{Ag}_{2} \mathrm{SO}_{3}\) is given below:

$$ \begin{aligned} \mathrm{K}_{\mathrm{IP}}=&\left[\mathrm{Ag}^{+}\right]_{\text {initial }}^{2}\left[\mathrm{SO}_{4}^{2-}\right]_{\text {initial }} \\ =&(0.0129)^{2}(0.0968) \\ &=1.6 \times 10^{-5} \end{aligned} $$

The ionic product \(\left(\mathrm{K}_{\mathrm{IP}}\right)\) is greater than the solubility product \(\left(\mathrm{K}_{\mathrm{sp}}\right)\), \(\mathrm{K}_{\mathrm{IP}}>\mathrm{K}_{\mathrm{sp}} \Rightarrow\) precipitationtakesplace

For the mixture, \(150.0 \mathrm{mLof} 0.10 \mathrm{MNa}_{2} \mathrm{SO}_{4}\) (aq) and 5.0mL of 0.40M AgNO \(_{3}\) (aq) precipitation takes place.

The concentration of silver ions \(\left[\mathrm{Ag}^{+}\right]\) and sulphate ions \(\left[\mathrm{SO}_{4}^{2-}\right]\) has been calculated to find the ionic product. The value of ionic product is greater than that of the solubility product; therefore, the precipitation does take place.

 

The balanced equation for the reaction of \(\mathrm{AgNO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) :

The initial concentration of \(\left[\mathrm{Ag}^{+}\right]\) ions in \(\mathrm{AgNO}_{3}\) is given below:

The initial concentration of \(\left[\mathrm{SO}_{4}^{2-}\right]\) ions in \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is given below:

$$ \begin{array}{c} {\left[\mathrm{SO}_{4}^{2-}\right]_{\text {initial }}=\frac{150.0 \mathrm{~mL} \times \frac{0.10 \mathrm{mmolNa}_{2} \mathrm{SO}_{4}}{\mathrm{~mL}} \times \frac{1 \mathrm{~mol} S O_{4}^{2-}}{1 \mathrm{molNa}_{2} \mathrm{SO}_{4}}}{(150.0+5.0) \mathrm{mL}}} \\ =0.0968 \mathrm{M} \end{array} $$

The ionic product of \(\mathrm{Ag}_{2} \mathrm{SO}_{3}\) is given below:

$$ \begin{aligned} \mathrm{K}_{\mathrm{IP}}=&\left[\mathrm{Ag}^{+}\right]_{\text {initial }}^{2}\left[\mathrm{SO}_{4}^{2-}\right]_{\text {initial }} \\ =&(0.0161)^{2}(0.0968) \\ &=2.5 \times 10^{-5} \end{aligned} $$

The ionic product \(\left(\mathrm{K}_{\mathrm{IP}}\right)\) is greater than the solubility product \(\left(\mathrm{K}_{\mathrm{sp}}\right)\), \(\mathrm{K}_{\mathrm{IP}}>\mathrm{K}_{\mathrm{sp}} \Rightarrow\) precipitationtakesplace

For the mixture, \(150.0 \mathrm{~mL}\) of \(0.10 \mathrm{MNa}_{2} \mathrm{SO}_{4}\) (aq) and 5.0mL of 0.50M AgNO \(_{3}\) (aq) precipitation takes place.

The balanced equation for the reaction of \(\mathrm{AgNO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is given in which the two moles of \(\mathrm{AgNO}_{3}\) react with one mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) to produce the \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) precipitate. The concentration of silver ions \(\left[\mathrm{Ag}^{+}\right]\) and sulphate ions \(\left[\mathrm{SO}_{4}^{2-}\right]\) has been calculated to find the ionic product. The value of ionic product is greater than that of the solubility product for which the precipitation does take place.