# What is the pH of a 0.10*mol*L^-1 solution of ammonium ion....for which pK_a=9.26?

Mar 31, 2017

And the $p {K}_{a}$ of ammonium ion is..............?

#### Explanation:

We interrogate the equilibrium,

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

For which, ${K}_{a} = \frac{\left[N {H}_{3}\right] \left[{H}_{3} {O}^{+}\right]}{\left[N {H}_{4}^{+}\right]} = {10}^{- 9.25}$

So ${10}^{- 9.25} = {x}^{2} / \left(0.10 - x\right)$, a quadratic in $x$, that we can solve by successive approximation, i.e. assuming that $0.1 > x$:

Thus $x \cong \sqrt{{10}^{- 9.25} \times 0.10}$

${x}_{1} = 7.5 \times {10}^{-} 6$

${x}_{2} = 7.5 \times {10}^{-} 6$

Since $x = \left[{H}_{3} {O}^{+}\right]$, $p H = - {\log}_{10} \left(7.5 \times {10}^{-} 6\right) = 5.12$