# What is pH of an aqueous solution for which [NH_4^+]=1.0*mol*L^-1?

Mar 23, 2017

It should be under $7$, should it not?

#### Explanation:

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$.

And we know that ammonium salts are weakly acidic, and undergo the acid-base reaction:

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

For ammonium ion, ${K}_{a} = 1.80 \times {10}^{-} 5$

And if there were a $1.0 \cdot m o l \cdot {L}^{-} 1$ solution of ammonium ion, we would solve the equilibrium expression in the usual way.

i.e. ${K}_{a} = 1.80 \times {10}^{-} 5 = \frac{\left[{H}_{3} {O}^{+}\right] \left[N {H}_{3} \left(a q\right)\right]}{\left[N {H}_{4}^{+}\right]}$

And if $x \cdot m o l \cdot {L}^{-} 1$ ammonium ion ionizes, then.........

${K}_{a} = 1.80 \times {10}^{-} 5 = {x}^{2} / \left(\left(1.0 - x\right) \cdot m o l \cdot {L}^{-} 1\right)$, a quadratic in $x$, which may be simplified given the reasonable assumption that $1.0 \text{>>} x$, and thus $1.0 - x \cong 1.0$.

And thus $x \cong \sqrt{{1.80}^{-} 5}$,

so ${x}_{1} = 4.2 \times {10}^{-} 3$, and recycling this approximation of $x$ back into the expression:

${x}_{2} = 4.2 \times {10}^{-} 3$, and since $x$ has converged, we may accept this value.

Thus $\left[{H}_{3} {O}^{+}\right] = 4.2 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$, and $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = - {\log}_{10} \left(4.2 \times {10}^{-} 3\right) = 2.37$.