# If the K_b of a weak base is 1.2 times 10^-6, what is the pH of a 0.34 M solution of this base?

May 4, 2017

It does not matter what base you are looking at. If you know its ${K}_{b}$ and its concentration, that is enough. Denote a neutral base as $B$. Then the ICE table can be constructed.

${\text{B"(aq) " "+" " "H"_2"O"(l) rightleftharpoons "BH"^(+)(aq) + "OH}}^{-} \left(a q\right)$

$\text{I"" ""0.34 M"" "" "" "-" "" "" ""0 M"" "" "" ""0 M}$
$\text{C"" "-x" "" "" "" "-" "" "+x" "" "" } + x$
$\text{E"" "(0.34 - x)"M"" "-" "" "" "x" "" "" "" } x$

Thus, the ${K}_{b}$ is:

${K}_{b} = {x}^{2} / \left(0.34 - x\right)$

In fact, the ${K}_{b}$ is small enough to use the small $x$ approximation. A good rule of thumb is to check whether $K$ is on the order of ${10}^{- 5}$. Hence, we have:

${K}_{b} \approx {x}^{2} / 0.34$

and the equilibrium concentration of ${\text{OH}}^{-}$ is readily obtained:

$x = \sqrt{0.34 {K}_{b}}$

$= 6.39 \times {10}^{- 4}$ $\text{M}$

Therefore, the $\text{pH}$ is:

color(blue)("pH") = 14 - "pOH"

$= 14 - \left(- \log \left[{\text{OH}}^{-}\right]\right)$

$= 14 - 3.19$

$= \textcolor{b l u e}{10.81}$