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 0.42mL of a solution containing 0.42g of sodium stearate?

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${\text{C}}_{17}{\text{H}}_{35}\text{COO}$willreact 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.

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

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