We are being asked to calculate the pH for a 0.250 mol NH_{3} in 1 L solution

NH_{3} is a **neutral amine**, therefore we know that it is a **weak base**.

NH_{3}_{(aq)} + H_{2}O_{(aq)} ⇌ OH^{–}_{(aq)} + NH_{4}^{+}_{(aq)}; **K _{b} = 1.8 × 10^{-5}**

Calculate the molarity:

$\overline{){\mathbf{Molarity}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{moles}\mathbf{}\mathbf{solute}}{\mathbf{L}\mathbf{}\mathbf{solution}}}\phantom{\rule{0ex}{0ex}}\mathbf{Molarity}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{}\mathbf{mol}}{\mathbf{1}\mathbf{}\mathbf{L}}$

**Molarity = 0.250 M**

From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table.

The ** K_{b} expression** for NH

$\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{products}}{\mathbf{reactants}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}\mathbf{\left[}{{\mathbf{NH}}_{\mathbf{4}}}^{\mathbf{+}}\mathbf{\right]}}{\mathbf{\left[}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\right]}}}$

Note that each concentration is raised by the stoichiometric coefficient: [NH_{3}], [OH^{–}] and [NH_{4}^{+}] are raised to 1.

What is the pH of a solution in which 0.250 mol of NH_{3} (K_{b} = 1.8 × 10^{-5}) is dissolved in sufficient water to make 1.00 L of solution?

a. 2.67

b. 11.33

c. 4.93

d. 13.4

e. 9.07

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Weak Bases concept. If you need more Weak Bases practice, you can also practice Weak Bases practice problems.