# A 100mL sample of 0.10 M NH3 has a Kb of 1.8*10^-5. What is the pH?

Nov 3, 2015

$\textcolor{b l u e}{p H = 11.13}$

#### Explanation:

In aqueous solution, $N {H}_{3}$ reacts with water according to the following reaction:

$\text{ " " " " " " } N {H}_{3} \left(a q\right) + {H}_{2} O \left(l\right) \to N {H}_{4}^{+} \left(a q\right) + O {H}^{-} \left(a q\right)$
$I n i t i a l \text{ " " "0.10M" " " " " " " " " " " "0M" " " " " } 0 M$
$\text{Change" " " " "-x" " " " " " " " " " " "+x" " " " " } + x$
$E q u i l i b r i u m \text{ "(0.10-x)M " " " " " " ""+x" " " " " } + x$

The equilibrium constant is written as:
${K}_{b} = \frac{\left[N {H}_{4}^{+}\right] \left[O {H}^{-}\right]}{N {H}_{3}} = 1.8 \times {10}^{- 5}$

Replacing the equilibrium concentrations by their values in the expression of ${K}_{b}$:

${K}_{b} = \frac{\left(x\right) \left(x\right)}{\left(0.10 - x\right)} = 1.8 \times {10}^{- 5}$

since the value of ${K}_{b}$ value is small, we consider $x \text{<<} 0.10$

Solving for $x$, $x = 1.34 \times {10}^{- 3} M$

$x$ represents the concentration of $O {H}^{-}$.
Using the expression of ${K}_{w} = \left[{H}^{+}\right] \left[O {H}^{-}\right] = 1.0 \times {10}^{- 14}$
$\left[{H}^{+}\right] = \frac{{K}_{w}}{\left[O {H}^{-}\right]} = \frac{1.0 \times {10}^{- 14}}{1.34 \times {10}^{- 3}} = 7.46 \times {10}^{- 12} M$

Then, the pH is calculated by:
$p H = - \log \left(\left[{H}^{+}\right]\right)$ $\implies p H = - \log \left(7.46 \times {10}^{- 12}\right)$

$\textcolor{b l u e}{p H = 11.13}$