# What is the concentration of ammonium ion in a solution of NH_3(aq) that is nominally 0.157*mol*L^-1?

Apr 10, 2017

$\left[N {H}_{4}^{+}\right] = 1.67 \times {10}^{-} 3 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We interrogate the equilibrium:

NH_3(aq) + H_2O(l) rightleftharpoonsNH_4^+ + ""^(-)OH,

for which ${K}_{b} = 1.8 \times {10}^{-} 5 = \frac{\left[N {H}_{4}^{+}\right] \left[H {O}^{-}\right]}{\left[N {H}_{3} \left(a q\right)\right]}$

Initially, $\left[N {H}_{3}\right] = 0.157 \cdot m o l \cdot {L}^{-} 1$, and we assume some ammonia accepts a proton from water to give stoichiometric $N {H}_{4}^{+}$ and $H {O}^{-}$. If we let the amount of protonation be $x$, then we write:

${K}_{b} = 1.8 \times {10}^{-} 5 = \frac{{x}^{2}}{0.157 - x}$

This is a quadratic in $x$, which we could solve exactly, but we make that approximation that $0.157 - x \cong 0.157$, and thus........

${x}_{1} \cong \sqrt{1.8 \times {10}^{-} 5 \times 0.157} = 1.68 \times {10}^{-} 3$

This is indeed small compared to $0.157$, but we could plug it in again, now that we have an approx. for $x$.

${x}_{2} = 1.67 \times {10}^{-} 3$, and since the values have converged, I am willing to accept this value for our answer.

Now $x = \left[N {H}_{4}^{+}\right] = \left[H {O}^{-}\right]$ by our definition. What is the $p H$ of this solution?