Question 5b2c0

Dec 1, 2015

$6.6 \cdot {10}^{- 3} \text{M}$

Explanation:

This is a pretty straightforward practice problem on using an ICE table to figure out the equilibrium concentration of the hydroxide ions.

Calculate the initial molarity of methylamine by using the number of moles and the volume of the solution

$\textcolor{b l u e}{c = \frac{n}{V}}$

$c = \text{0.10 moles"/"1.0 L" = "0.10 M}$

Before moving on to the actual ICE table, take a look at the value of the base dissociation constant, ${K}_{b}$.

When ${K}_{b} < 1$, the reaction will favor the reactants. This means that you can expect the equilibrium concentration of the two products to be significantly smaller than the equilibrium constant of the reactant.

So, an ICE table for this reaction would look like this

"CH"_3"NH"_text(2(aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "CH"_3"NH"_text(3(aq])^(+) " "+" " "OH"_text(aq])^(-)

color(purple)("I")" " " "0.10" " " " " " " " " " " " " " " " " " " " " " "0" " " " " " " " " " " " " "0
color(purple)("C")" "(-x)" " " " " " " " " " " " " " " " " " " "(+x)" " " " " " " "(+x)
color(purple)("E")" "0.10-x" " " " " " " " " " " " " " " " " " " "x" " " " " " " " " " " "x#

By definition, the base dissociation constant will be equal to

${K}_{b} = \left(\left[{\text{OH"^(-)] * ["CH"_3"NH"_3^(+)])/(["CH"_3"NH}}_{3}\right]\right)$

In your case, this will be equal to

${K}_{b} = \frac{x \cdot x}{0.10 - x} = {x}^{2} / \left(0.10 - x\right)$

Because ${K}_{b}$ is so small, you can use the approximation

$0.10 - x \approx 0.10$

This means that you have

${K}_{b} = {x}^{2} / 0.10 = 4.4 \cdot {10}^{- 4}$

This will get you

$x = \sqrt{0.10 \cdot 4.4 \cdot {10}^{- 4}} = 6.6 \cdot {10}^{- 3}$

Since $x$ represents the equilibrium concentration of the hydroxide anions, you will have

$\left[\text{OH"^(-)] = color(green)(6.6 * 10^(-3)"M}\right)$