Q18.186CP - Chemistry: Molecular Nature Of Matter And Change

Question

A solution of propanoic acid $\left({\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}\right)$,made by dissolving $\mathbf{7}\mathbf{.500}\mathbf{ }\mathbf{\text{g}}$ in sufficient water to make $\mathbf{100}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{mL}}$, has a freezing point of $\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{890}{\mathbf{ }}^{\mathbf{\text{0}}}\mathbf{\text{C}}$. 

(a) Calculate the molarity of the solution. 

(b) Calculate the molarity of the propanoate ion. (Assume the molarity of the solution equals the molality.) 

(c) Calculate the percent dissociation of propanoic acid

Step-by-Step Solution

Answer

1. The molarity of the solution $1.012 \text{M}$
2. The molarity of the propanoate ion is $4\times {10}^{-3} \text{M}$
3. The percent dissociation of propanoic acid is, $0.395\%$

Step 1: To Calculate the molarity of the solution

(a)

Solve the concentration of the solution using the given mass and volume.

$$\begin{gathered} M=\frac{mol}{L} \\ =\frac{7.500 \text{g}\times \frac{1 \text{mol}}{74.08 \text{g}}}{100 \text{mL}\times \frac{1 \text{L}}{1000 \text{mL}}} \\ = 1.012 \text{M} \end{gathered}$$

Hence the molarity of the solution is, $1.012 \text{M}$

Step 2 : To Calculate the molarity of the propanoate ion

(b)

Write the equation of the dissociation of propanoic acid.

${\text{CH}}_3{\text{CH}}_2\text{COOH}+{\text{H}}_2\text{O}\rightleftharpoons {\text{CH}}_3{\text{CH}}_2{\text{COO}}^{-}+{\text{H}}_3{\text{O}}^{+}$

Next, construct the ICE table to obtain the equation for ${\text{K}}_{\text{a}}$

$\begin{array}{ccccc}\mathrm{I}& \left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}\mid {\text{ + H}}_{\text{2}}\text{O}\right.& \mathit{\text{C}}& \left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{ - }}\right]\text{ + }& \left.\mid {\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\\ \text{C}& \text{1.012M}& 0& 0& \\ \text{F}& \text{ - x}& & \text{ + x}& \text{ + x}\end{array}$

Solve for the molality using the freezing point depression equation.

$$\begin{gathered} \Delta T_f=K_f \text{mm } \\ =\frac{\Delta T_f}{K_f} \\ =\frac{0.00 {}^{\text{0}}\text{C}-\left(-1.890 {}^{\text{0}}\text{C}\right)}{\left(1.86 {}^{\text{0}}\text{C/m}\right)} \text{m } \\ =1.016 \end{gathered}$$

Note that the molality is just equal to molarity. The total concentration of the solution is 1.016. Solve for data-custom-editor="chemistry" $\left[{\text{CH}}_3{\text{CH}}_2{\text{COO}}^{-}\right]$ which is equal to x in the ice table.

data-custom-editor="chemistry" $$\begin{gathered} \text{ Total conc = }\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}\right]\text{ + }\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{COO}}^{\text{ - }}\right]\text{ + }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\mathrm{l} \\ 1.016\text{M }=1.012\text{M- x + x + x} \\ \text{x} =1.016\text{M }-1.012\text{M} \\ \text{x }=0.004\text{M} \end{gathered}$$

Since $x = \left[{\text{CH}}_3{\mathrm{CH}}_2{\mathrm{COO}}^{-}\right],\text{then}\left[{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{COO}}^{-}\right]= 4\times {10}^{-3} \text{M}$

Hence the molarity of the propanoate ion is $4\times {10}^{-3} \text{M}$ 

Step 3: To Calculate the percent dissociation of propanoic acid

(c)

The formula for percent dissociation is

$\text{\% dissociation}=\frac{{\text{[HA]}}_{\text{dissoc }}}{{\text{[HA]}}_{\text{int }}}\times 100\%$

Use this to solve.

$\%\text{ dissociation }=\frac{{\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}\right]}_{\text{dissoc }}}{{\left[{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{COOH}\right]}_{\text{int }}}\times 100\%$

$\%\text{dissociation}=0.395\%$

Hence the percent dissociation of propanoic acid is, $0.395\%$