Q18.174CP - Chemistry: Molecular Nature Of Matter And Change

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. 

Step-by-Step Solution

Answer

a) The solution is basic.

b) pH=8.93

Step 1: 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.

Step 2: Calculate the concentration

(b)

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

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

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

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.

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

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

Step 3: 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 ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{ - }}{\text{ is a weak base, the K}}_{\text{b}}$  must be very small. So, assume that the x has no effect on  in 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.