Chapter 18, Problem 18.172P

(a)

Interpretation Introduction

Interpretation:

The aqueous calcium propionate has to be identified whether the solution is acidic, basic or neutral.

Concept introduction:

The aqueous of acidic, basic or neutral is identified by using following method.

The depending on an ion’s ability to react with water,

If the solution is neutral, the anion from strong acid and the cation from strong base.

If the solution is acidic, the anion from strong acid (or highly charged metal cation) and the cation from weak base.

If the solution is basic, the anion from weak acid and the cation from strong base.

Acidic solution:

If the solution is acidic, I would form from cation of weak base and anion of strong acid.

In addition, small, highly charged metal cation and anion of strong acid are acidic solution.

The $\text{pH}$ of the solution is less than seven.

Basic solution:

If the solution is basic, I would form from cation of strong base and anion of weak acid.

Neutral solution:

If the solution is neutral, I would form from cation of strong base and anion of strong acid.

Explanation of Solution

The given compound is shown below,

${\text{Calcium propionate Ca(CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO)}}_{\text{2}}\text{; calcium propanoate}\text{.}$

The equation is shown below,

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}\text{(aq)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(l) }\rightleftharpoons {\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH(aq)+}{\text{OH}}^{\text{-}}\text{(aq)}$

The two ions are produced from the strong calcium hydroxide, calcium hydroxide is strong base, therefore it won’t influence the $\text{pH}$ of the solution. The ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}$ formed from the conjugate base of weak propionic acid, so it will act as weak base in solution, therefore it is basic in pure water.

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution has to be calculated for $\text{8}\text{.75 g of Ca}{\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COO}\right)}_{\text{2}}$ dissolves in water to give $\text{0}\text{.500 L}$ of solution.

Concept introduction:

An equilibrium constant $\left(K\right)$ is the ratio of concentration of products and reactants raised to appropriate stoichiometric coefficient at equlibrium.

For the general base B,

  $\text{B}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{BH}}^{\text{+}}\left(\text{aq}\right){\text{+OH}}^{\text{-}}\left(\text{aq}\right)$

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

  ${\text{K}}_{\text{b}}\text{=}\frac{\left[{\text{BH}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]}{\left[\text{B}\right]}$                                                                                                 $\left(1\right)$

An equilibrium constant $\left(K\right)$ with subscript b indicate that it is an equilibrium constant of an base in water.

$\begin{array}{l}{\text{Base - dissociation constants can be expressed as pK}}_{\text{b}}\text{ values,}\\ {\text{pK}}_{\text{b}}{\text{ = -log K}}_{\text{b}}\text{ and}\\ {\text{10}}^{{\text{ - pK}}_{\text{b}}}{\text{ = K}}_{\text{b }}\end{array}$

Percent dissociation can be calculated by using following formula,

  $\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}$

The ${\text{K}}_{\text{b}}$ value is calculating by using following formula,

  ${\text{K}}_{\text{w}}{\text{ = K}}_{\text{a}}{\text{ × K}}_{\text{b}}$

The molarity is calculated by using following formula,

  $\text{Molarity}\text{(M)}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{liter}}$

Explanation of Solution

Given,

The $\text{pH}$ of the solution (vinegar with $\text{5}\text{.0\% }\left(\text{w/v}\right)$ acetic acid) depends on the molal concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$.

Calcium propionate dissolves in water and gives propionates ion.

  ${\text{Ca(CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO)}}_{\text{2}}\text{(aq)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(l) }\rightleftharpoons 2{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}\text{(aq)+}{\text{Ca}}^{\text{2+}}\text{(aq)}$

The molarity is calculated by using following formula,

  $\begin{array}{l}\text{Molarity}\text{(M)}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{liter}}\\ \text{Molarity}\text{(M)}\text{=}\frac{\text{8}\text{.75}\text{g}{\text{Ca(CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO)}}_{\text{2}}}{\text{1}\times {\text{10}}^{-3}L}\text{×}\frac{\text{1 mol}{\text{Ca(CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO)}}_{\text{2}}}{\text{186}\text{.22}\text{g}{\text{Ca(CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO)}}_{\text{2}}}\text{×}\frac{2\text{mol}{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}}{\text{1}\text{mol}{\text{Ca(CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO)}}_{\text{2}}}\\ \text{Molarity}\text{(M)}\text{= 0}\text{.18794}M{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}\end{array}$

The balance equation is given below,

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}\text{(aq)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(l) }\rightleftharpoons {\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH(aq)+}{\text{OH}}^{\text{-}}\text{(aq)}$

$\text{0}\text{.18794}M{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}$ Solution.

Therefore,

ICE table:

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}\text{(aq)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(l) }\rightleftharpoons {\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH(aq)+}{\text{OH}}^{\text{-}}\text{(aq)}$

Initial concentration	0.1879 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.1879 -x		x	x

The initial concentration is $\text{0}\text{.18794}M{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}$.

  $\begin{array}{l}{\text{K}}_{\text{w}}{\text{= K}}_a\times {\text{K}}_b\\ {\text{K}}_b\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_a}\\ {\text{K}}_b\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{1.3\times {10}^{-5}}\\ {\text{K}}_b\text{= 7}{\text{.69×10}}^{\text{-10}}\end{array}$

  $\begin{array}{l}{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}\text{(aq)}\text{+}{\text{H}}_{\text{2}}\text{O}\text{(l) }\rightleftharpoons {\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH(aq)+}{\text{OH}}^{\text{-}}\text{(aq)}\\ The value{\text{ K}}_{\text{b}}of{\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_b\text{ =}\frac{\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}\right]}\\ \text{given,}\\ {\text{K}}_b\text{ =}\frac{\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}\right]}=7.69\times {10}^{-10},\\ letconsider,\\ 0.1879-x=0.1879\\ \frac{\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}\right]}=7.69\times {10}^{-10},\\ \frac{x^2}{(0.1879)}=7.69\times {10}^{-10}\\ x^2\text{=}1.44\times {10}^{-10}\\ \text{x}\text{=}1.20\times {10}^{-5}M=\left[O{H}^{-}\right]\end{array}$

  $\begin{array}{l}\text{x is small when compared to 0}\text{.1879}\text{M, therefore,}\\ \text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \frac{1.20\times {10}^{-5}}{\text{0}\text{.1879}\text{M}}\text{×100=}0.006\\ \text{The}\text{dissociation}\text{value}\text{is}\text{very}\text{less, and}\text{it }\text{is }\text{valid}\\ {\text{K}}_{\text{w}}\text{=}\left[{\text{OH}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{1.20\times {10}^{-5}}\\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{= 8}{\text{.33×10}}^{\text{-10}}\\ \text{Therefore,}\\ \text{pH}\text{=}\text{-}\text{log}{\text{(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\\ \text{pH}\text{=}\text{-}\text{log}\text{(8}{\text{.33×10}}^{\text{-10}}\text{)}\\ \text{pH}\text{=}9.07\end{array}$

 $\text{pH}$ of the solution is 9.1.