Question

Calcium propionate$\mathbf{\text{Ca}}({\text{CH}}_3{\text{CH}}_2\text{COO}{\mathbf{)}}_{\mathbf{2}}$; calcium propionate] is a mold inhibitor used in food, tobacco, and pharmaceuticals.

(a) Use balanced equations to show whether aqueous calcium propionate is acidic, basic, or neutral.

(b) Use Appendix C to find the resulting pH when 8.75g of$\mathbf{\text{Ca}}({\text{CH}}_3{\text{CH}}_2\text{COO}{\mathbf{)}}_{\mathbf{2}}$dissolves in enough water to give 0.500L of solution.

Short answer

a) The solution is basic.

b) pH=8.93

Step-by-step solution

Show whether the aqueous calcium propionate is acidic, basic, or neutral

(a)

$\text{Ca}{\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COO}\right)}_{\text{2}}$is a salt and will dissolve in water

$\text{Ca}{\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COO}\right)}_2\to {\text{Ca}}^{\text{2 +}}\text{+ 2}{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}$

Then${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}$reacts with water.

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

Since the reaction of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{produces OH}}^{\text{-}},$the solution is basic.

Calculate the concentration

(b)

First, calculate the concentration of$\text{Ca}{\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COO}\right)}_2$.

$$\begin{gathered} \text{M}\text{=}\frac{\text{mol}}{\text{L}} \\ =\frac{8.75g\times {\displaystyle \frac{1mol}{186.22g}}}{0.500L} \\ =\text{0.094M.} \end{gathered}$$

$\text{Ca}{\left({\text{CH}}_3{\text{CH}}_2\text{COO}\right)}_2$

Salt will dissolve in water and produce ions with the same concentration in the reaction written in a. Then, the ${\mathbf{\text{CH}}}_{\mathbf{3}}{\mathbf{\text{CH}}}_{\mathbf{\text{3}}}$will react with water to produce a basic solution.

${\text{CH}}_{\text{3}}{\text{C}}_{\text{2}}{\text{C}}^Y{\text{OO}}^{-}{\text{+11}}_2{\text{O⇌C}}^Y{\text{11}}_3{\text{C}}_{\text{1}}{\text{C}}^Y{\text{O)H + OII}}^{-}$

$$\begin{gathered} {\text{K}}_{\text{b}}\text{=}\frac{\left.\left[{\text{OH}}^{\text{-}}\right]{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}\right]}{\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}\right]} \\ \text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.094 - x}}. \end{gathered}$$

Find the resulting pH

Solve for the${\text{K}}_{\text{b}}$from the${\text{K}}_{\text{a}}$of the propanoic acid from Appendix C.

$$\begin{gathered} {\text{K}}_{\text{w}}\text{=}{\text{K}}_{\text{a}}{\text{×K}}_{\text{b}} \\ {\text{K}}_{\text{b}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}} \\ \text{=}\frac{1\times {10}^{-14}}{1.3\times {10}^{-5}} \\ {\text{=7.69×10}}^{-10} \end{gathered}$$

We know that$\text{x}\text{=}\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]$. Since${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{is a weak base, theK}}_{\text{b}}$ must be very small. So, assume that the x has no effect onin the denominator. Then replace the${\text{K}}_{\text{b}}$to solve for the x.

$$\begin{gathered} {\text{K}}_{\text{b}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0.094}} \\ {\text{x}}^{\text{2}}\text{=}\left({\text{K}}_{\text{b}}\right)\text{(0.094)} \\ \text{x =}\sqrt{\left({\text{K}}_{\text{b}}\right)\text{(0.094)}} \\ =\sqrt{(7.69\times {10}^{-10})(0.094)} \\ ={\text{8.50×10}}^{-6} \end{gathered}$$

Since $\text{x}\text{=}\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]\text{,then}\left[{\text{OH}}^{\text{-}}\right]\text{=}{\text{8.50×10}}^{-6}\text{M.}$

Solve for the pOH.

$$\begin{gathered} \text{pOH}\text{=}\text{- log}\left[{\text{OH}}^{\text{-}}\right] \\ \text{=}{\text{- log(8.50×10}}^{-6}\text{)} \\ =5.07 \end{gathered}$$

Lastly, solve for the pH.

$$\begin{gathered} \text{pH + pOH}\text{=}\text{14pH} \\ \text{pH}\text{=}\text{14 - pOH} \\ \text{= 14 - 5.07} \\ \text{=}\text{8.93.} \end{gathered}$$

Hence, the pH=8.93.