# If "0.50 M" of "NH"_4"OH" were to dissociate in water, and K_b = 1.773 xx 10^(-5), what is the resultant "pH" at equilibrium?

Jun 8, 2017

$p H = 11.47$

#### Explanation:

Let's make an (R)ICE table for the dissociation of $N {H}_{4} O H$ (the base you get when you dissolve $N {H}_{3}$ in water.)

$N {H}_{3} + {H}_{2} O r i g h t \le f t h a r p \infty n s N {H}_{4} O H$

Suppose that $x$ $M$ of $N {H}_{4} O H$ then dissociated. ($x \ge 0$)

Reaction: $N {H}_{4} O H \text{ "rightleftharpoons" } N {H}_{4}^{+} + O {H}^{-}$

Initial: $\text{ "" ""0.50 M}$$\text{ "" "" }$$\text{0 M}$$\text{ "" "" }$$\text{0 M}$

Change: $\text{ "-x" M}$$\text{ }$$\text{ } + x$ $\text{M"" "" }$ $+ x$ $\text{M}$

Equil.: $\text{ } \left(0.50 - x\right)$ $\text{M}$$\text{ }$$x$ $\text{M"" "" }$ $\text{ } x$ $\text{M}$

Using the mass action expression for ${K}_{b}$:

${K}_{b} = \frac{\left[N {H}_{4}^{+}\right] \left[O {H}^{-}\right]}{\left[N {H}_{4} O H\right]} = \frac{{x}^{2}}{0.50 - x}$

Solve for $x$:

given that ${K}_{b} = 1.773 \cdot {10}^{-} 5$,

$x = 0.002969$ $M$

Thus $\left[O {H}^{-}\right] = 0.002969$ $M$

Evaluate the $p H$:

$\textcolor{b l u e}{p H} = 14 - p O H = 14 + \log \left[O {H}^{-}\right] = \textcolor{b l u e}{11.47}$

to two sig figs. :)