Chapter 17, Problem 17.108QP

(a)

Interpretation Introduction

Interpretation:

The molar solubility, $\text{pH}$ and concentration of given solution has to be calculated.

Concept introduction:

• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base. In general, addition of acid or base does not affect the $\text{pH}$ in buffer solution but if it is more than amount of conjugate base or conjugate acid, then buffer loses its buffering capacity.
• Buffer solution is a combination of a weak acid and its conjugate base or vice-versa in an aqueous solution.
• Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
• When the concentration of one of the ions of a chemical solution got higher, it reacts with counter charged ions and precipitated out as salt till the ion product equals solubility product is called common ion effect.

To calculate: the molar solubility of ${\text{CaCO}}_{\text{3}}$ in ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$

Answer to Problem 17.108QP

$\text{s = 1}\text{.7}{\text{×10}}^{\text{-}}{}^{\text{7}}\text{M}$

Explanation of Solution

  $\begin{array}{l}\text{The}\text{dissociation}\text{of}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\\ \\ {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\left(\text{s}\right)}\stackrel{{\text{H}}_{\text{2}}\text{O}}{\to }{\text{2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{ + CO}}_{\text{3}}{{}^{\text{2}}}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{2}\left(\text{0}\text{.050M}\right)\text{ 0}\text{.050M}\\ \\ {\text{Consider s be the molar solubility of CaCO}}_{\text{3}}\text{in}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\\ \\ {\text{CaCO}}_{\text{3}}{}_{\left(\text{s}\right)}\stackrel{{\text{H}}_{\text{2}}\text{O}}{\to }{\text{Ca}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{ + CO}}_{\text{3}}{{}^{\text{2}}}^{\text{-}}{}_{\left(\text{aq}\right)}\\ \text{Initial concentration}\left(\text{M}\right)\text{: 0}\text{.00 0}\text{.050 }\\ \text{Change}\text{in}\text{concentration }\left(\text{M}\right)\text{: +s +s }\\ \text{Equilibrium}\text{concentration }\left(\text{M}\right)\text{: } \text{+s 0}\text{.050+s}\\ \\ \begin{array}{l}{\text{K}}_{\text{sp}}{\text{ = [Ca}}^{\text{2+}}{\text{][CO}}_{\text{3}}{{}^{\text{2}}}^{\text{-}}\text{]}\hfill \\ \hfill \\ \text{8}\text{.7}\text{×}{\text{10}}^{\text{-}}{}^{\text{9}}\text{ = s}\left(\text{0}\text{.050+s}\right)\hfill \\ \hfill \\ \begin{array}{l}\text{Small and neglect it, }\\ \\ \text{0}\text{.050+s}\approx \text{0}\text{.050}\end{array}\hfill \\ \hfill \\ \text{8}\text{.7}\text{×}{\text{10}}^{\text{-}}{}^{\text{9}}\text{ = 0}\text{.050s}\hfill \\ \hfill \\ \text{s = 1}\text{.7}{\text{×10}}^{\text{-}}{}^{\text{7}}\text{M}\hfill \end{array}\end{array}$

The molar solubility of ${\text{CaCO}}_{\text{3}}$ in ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated using the solubility product expression of ${\text{CaCO}}_{\text{3}}$.  By substituting the concentrations of ions and doing simple mathematical operations, the molar solubility of ${\text{CaCO}}_{\text{3}}$ is determined.

(b)

Interpretation Introduction

Interpretation:

The molar solubility, $\text{pH}$ and concentration of given solution has to be calculated.

Concept introduction:

• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base. In general, addition of acid or base does not affect the $\text{pH}$ in buffer solution but if it is more than amount of conjugate base or conjugate acid, then buffer loses its buffering capacity.
• Buffer solution is a combination of a weak acid and its conjugate base or vice-versa in an aqueous solution.
• Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
• When the concentration of one of the ions of a chemical solution got higher, it reacts with counter charged ions and precipitated out as salt till the ion product equals solubility product is called common ion effect.

To explain: the reason of magnesium ion cannot be removed as above.

Answer to Problem 17.108QP

Because  are moderately soluble ${\text{(K}}_{\text{sp}}\text{=4}\text{.0}{\text{×10}}^{\text{-5}}\text{)}\text{.}$

Explanation of Solution

Because ${\text{Mg}}^{\text{2+}}\text{ions}$ are moderately soluble ${\text{(K}}_{\text{sp}}\text{=4}\text{.0}{\text{×10}}^{\text{-5}}\text{)}\text{.}$  Hence, it cannot be removed.

(c)

Interpretation Introduction

Interpretation:

The molar solubility, $\text{pH}$ and concentration of given solution has to be calculated.

Concept introduction:

• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base. In general, addition of acid or base does not affect the $\text{pH}$ in buffer solution but if it is more than amount of conjugate base or conjugate acid, then buffer loses its buffering capacity.
• Buffer solution is a combination of a weak acid and its conjugate base or vice-versa in an aqueous solution.
• Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
• When the concentration of one of the ions of a chemical solution got higher, it reacts with counter charged ions and precipitated out as salt till the ion product equals solubility product is called common ion effect.

To calculate the $\text{pH}$ of ${\text{Ca(OH)}}_{\text{2}}$

Answer to Problem 17.108QP

$\text{pH = 12}\text{.40}$

Explanation of Solution

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{of}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\text{is 8}\text{.0}\text{×}{\text{10}}^{\text{-6}}\\ \begin{array}{l}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\stackrel{}{\to }{\text{ Ca}}^{\text{2}}{}^{\text{+}}\text{+}{\text{2OH}}^{\text{-}}\hfill \\ \begin{array}{l}\text{At equilibrium: }\text{s 2s}\\ \\ \begin{array}{l} {\text{K}}_{\text{sp}} = 8.0\times {10}^{-}{}^6 = [{\text{Ca}}^{\text{2}}{}^{\text{+}}{\text{][OH}}^{\text{-}}{]}^2\hfill \\ \hfill \\ {\text{4s}}^{\text{3}}\text{ = 8}\text{.0}\text{×}{\text{10}}^{\text{-}}{}^{\text{6}}\hfill \\ \hfill \\ \begin{array}{l} \text{s = 0}\text{.0126M}\\ \\ \begin{array}{l} [{\text{OH}}^{\text{-}}] =\text{ 2s = 0}\text{.0252M}\hfill \\ \hfill \\ \text{pOH = -log}\left(\text{0}\text{.0252}\right)\text{ = 1}\text{.60}\hfill \\ \hfill \\ pH = 12.40\hfill \end{array}\end{array}\hfill \end{array}\end{array}\hfill \end{array}\end{array}$

The molar solubility of ${\text{Ca(OH)}}_{\text{2}}$ is calculated using the solubility product expression of .  By calculating the $\text{pOH}$ using concentration of hydroxide ion, the $\text{pH}$ value is determined.

(d)

Interpretation Introduction

Interpretation:

The molar solubility, $\text{pH}$ and concentration of given solution has to be calculated.

Concept introduction:

• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base. In general, addition of acid or base does not affect the $\text{pH}$ in buffer solution but if it is more than amount of conjugate base or conjugate acid, then buffer loses its buffering capacity.
• Buffer solution is a combination of a weak acid and its conjugate base or vice-versa in an aqueous solution.
• Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
• When the concentration of one of the ions of a chemical solution got higher, it reacts with counter charged ions and precipitated out as salt till the ion product equals solubility product is called common ion effect.

To calculate the strength of ${\text{Mg}}^{\text{2+}}$ ions

Answer to Problem 17.108QP

$[{\text{Mg}}^{\text{2}}{}^{\text{+}}\text{] = 1}\text{.9}\times {\text{10}}^{\text{-}}{}^{\text{8}}\text{M}$

Explanation of Solution

Higher concentration of hydroxide ion removes most of ${\text{Mg(OH)}}_{\text{2}}$.  But there is only small amount remaining due to below equilibrium

  $\begin{array}{l} \text{Mg}{\left(\text{OH}\right)}_{\text{2}}{}_{\left(\text{s}\right)}\stackrel{}{\to }{\text{ Mg}}^{\text{2}}{{}^{\text{+}}}_{\left(\text{aq}\right)}{\text{+2OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\hfill \\ \hfill \\ {\text{K}}_{\text{sp}}{\text{ = [Mg}}^{\text{2}}{}^{\text{+}}{\text{][OH}}^{\text{-}}{\text{]}}^{\text{2}}\hfill \\ \hfill \\ 1.2\times {10}^{-}{}^{11} = [{\text{Mg}}^{\text{2}}{}^{\text{+}}]{\left(0.0252\right)}^2\hfill \\ \hfill \\ [{\text{Mg}}^{\text{2}}{}^{\text{+}}\text{] = 1}\text{.9}\times {\text{10}}^{\text{-}}{}^{\text{8}}\text{M}\hfill \end{array}$

The concentration of ${\text{Mg}}^{\text{2+}}$ ions is calculated using the solubility product expression of ${\text{Mg(OH)}}_{\text{2}}$.  By substituting the concentrations of ion and ${\text{K}}_{\text{sp}}$ and doing simple mathematical operations, concentration of ${\text{Mg}}^{\text{2+}}$ ions is determined.

(e)

Interpretation Introduction

Interpretation:

The molar solubility, $\text{pH}$ and concentration of given solution has to be calculated.

Concept introduction:

• Buffer solution is defined as a solution that oppose changes in $\text{pH}$ while adding little amount of either an acid or a base.   In general, addition of acid or base does not affect the $\text{pH}$ in buffer solution but if it is more than amount of conjugate base or conjugate acid, then buffer loses its buffering capacity.
• Buffer solution is a combination of a weak acid and its conjugate base or vice-versa in an aqueous solution.
• Molar solubility is defined as amount of solute that can be dissolved in one litre of solution before it attains saturation.
• When the concentration of one of the ions of a chemical solution got higher, it reacts with counter charged ions and precipitated out as salt till the ion product equals solubility product is called common ion effect.

To explain: which one ions can be removed first.

Answer to Problem 17.108QP

Because calcium ion is present in huge amount

Explanation of Solution

As calcium ion is present in huge amount.  It can be removed first.