Q173CP - Chemistry: Molecular Nature Of Matter And Change

Question

Sodium stearate $\left({\text{C}}_{\text{17}}{\text{H}}_{\text{35}}\text{COONa}\right)$ is a major component of bar soap. The Ka of the stearic acid is $\mathbf{\text{1.3}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{\text{ - 5}}}$. What is the pH of 0.42  mL of a solution containing 0.42 g of sodium stearate?

Step-by-Step Solution

Answer

$\text{pH} \text{ = } \text{9.01.}$

Step 1: Calculate the concentration of sodium stearate in the solution

First, calculate the concentration of sodium stearate in the solution.

$\begin{array}{rcl}& & \text{M} \text{ = } \frac{\text{mol}}{\text{L}} \\ & =& \frac{0.42 \mathrm{g}\times {\displaystyle \frac{1 \mathrm{mol}}{306.5 \mathrm{g}}}}{10 \mathrm{mL} {\displaystyle \frac{1 \mathrm{L}}{1000 \mathrm{mL}}}}\\ & =& 0.137 \mathrm{M}\end{array}$

Then, write the reaction equations.

First, the sodium stearate will dissolve in water.

${\text{C}}_{\text{17}}{\text{H}}_{\text{35}}\text{COONa}\to {\text{C}}_{\text{17}}{\text{H}}_{\text{35}}{\text{COO + Na}}^{\text{ + }}$

Then, the ${\text{C}}_{17}{\text{H}}_{35}\text{COO}$  will react with water.

${\text{C}}_{\text{17}}{\text{H}}_{\text{35}}{\text{COO + H}}_{\text{2}}\text{O}\rightleftharpoons {\text{C}}_{\text{17}}{\text{H}}_{\text{35}}\text{COOH + OH}$

Now, solve for the ${\text{K}}_{\text{b}}$  using ${\text{K}}_w$ and the given ${\text{K}}_{\alpha }$.

$$\begin{gathered} {\text{K}}_{\text{w}}\text{ = } {\text{K}}_{\text{b}}\times {\text{K}}_{\text{a}} \\ {\text{K}}_{\text{b}}\text{ = } \frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\text{ = } \frac{\text{1.0}\times {\text{10}}^{\text{ - 14}}}{\text{1.3}\times {\text{10}}^{\text{ - 5}}} \\ \text{ = } \text{7.69}\times {\text{10}}^{\text{ - 10}}. \end{gathered}$$

Next, construct the ICE table to obtain the equation for K_b.

$$\begin{gathered} {\text{K}}_{\text{b}}\text{ = }\frac{\left[{\text{OH}}^{\text{ - }}\right]\left[{\text{C}}_{\text{17}}{\text{H}}_{\text{35}}\text{COOH}\right]}{\left[{\text{C}}_{\text{17}}{\text{H}}_{\text{35}}{\text{COO}}^{\text{ - }}\right]} \\ \text{ = }\frac{{\text{x}}^{\text{2}}}{\text{0.137 - x}}. \end{gathered}$$

Step 2: Find the pH

Since ${\text{C}}_{\text{17}}{\text{H}}_{\text{35}}{\text{COO}}^{\text{ - }}$is a weak base, its ${\text{K}}_{\text{b}}$must be very small. So, assume that the x has no effect on the 0.137 M  in the denominator. Then replace the ${\text{K}}_{\text{b}}$ to solve for x.

$\begin{array}{rcl}& & {\text{K}}_{\text{b}}\text{ = }\frac{{\text{x}}^{\text{2}}}{\text{0.137}}\\ & & {\text{x}}^{\text{2}}\text{ = }\left({\text{K}}_{\text{b}}\right)\text{(0.137)}\\ & & \text{x = }\sqrt{\left({\text{K}}_{\text{b}}\right)\text{(0.137)}}\\ & & \text{ = } \sqrt{\left(\text{7.69}\times {\text{10}}^{\text{ - 10}}\right)\text{(0.137)}}\\ & & \text{ = } \text{1.03}\times {\text{10}}^{\text{ - 5}}.\end{array}$

Since $\text{x} \text{ = } \left[{\text{OH}}^{\text{ - }}\right] \text{ = } \left[{\text{C}}_{\text{17}}{\text{H}}_{\text{35}}\text{COOH}\right]$, then $\left[{\text{OH}}^{\text{ - }}\right] \text{ = } \text{1.03}\times {\text{10}}^{\text{ - 5}} \text{M}$

Next, calculate the pOH of the solution.

$\begin{array}{rcl}& & \text{pOH} \text{ = } \text{ - log}\left[{\text{OH}}^{\text{ - }}\right]\\ & & \text{ = } \text{ - log}\left(\text{1.03}\times {\text{10}}^{\text{ - 5}}\right)\\ & & \text{ = } \text{4.99.}\end{array}$

Lastly, solve for the pH.

$\begin{array}{rcl}& & \text{pH + pOH = } \text{14}\\ & & \text{ pH = } \text{14 - pOH}\\ & & \text{ = } \text{14 - 4.99}\\ & & \text{ = } \text{9.01.}\end{array}$

Hence, the $\text{pH} \text{ = } \text{9.01.}$