# If the "pH" of an aqueous solution of ammonium chloride is 4.60, what is the K_h value of "NH"_4^(+)?

Mar 31, 2017

I assume it's an error in your notes? The relationship between ${K}_{a}$, ${K}_{b}$, and ${K}_{w}$ is that ${K}_{a} \cdot {K}_{b} = {K}_{w} = {10}^{- 14}$.

Even though you are not given ${K}_{a}$, you can still look it up. I happen to remember the ${K}_{b}$ for ${\text{NH}}_{3}$ though, so here's an opportunity to demonstrate the relationship.

The ${K}_{b}$ for ammonia, ${\text{NH}}_{3}$, is related to the ${K}_{a}$ for ammonium, ${\text{NH}}_{4}^{+}$.

${K}_{b} = 1.8 \times {10}^{- 5}$

${K}_{w} = {10}^{- 14}$

Therefore:

${K}_{a} = {K}_{w} / {K}_{b} = {10}^{- 14} / \left(1.8 \times {10}^{- 5}\right) = 5.55 \times {10}^{- 10}$

So, the ${K}_{a}$ for ${\text{NH}}_{4}^{+}$ is $5.55 \times {10}^{- 10}$, and this is what you would expect for your answer.. You should also be able to look this up in the Appendix of your textbook.

Now, pretend you didn't know that. Once you construct your ICE table:

${\text{NH"_4^(+)(aq) + "H"_2"O"(l) rightleftharpoons "NH"_3(aq) + "H"_3"O}}^{+} \left(a q\right)$

You should get:

K_a = (x^2)/(0.1 - x) = ?

If you aren't sure how I got here, tell me. ${\text{NH}}_{4}^{+}$ is a weak acid since ${\text{NH}}_{3}$ is a weak base. When you look at the ${K}_{b}$ for a weak base, it is moderately small. Therefore, for the conjugate weak acid, it is even smaller.

When we calculate ${K}_{a}$, do NOT use the small $x$ approximation, because it will not work unless ${K}_{a}$ is small enough. Beyond that, you should only use it when you use ${K}_{a}$ in the first place.

We know "pH" = 4.60 = -log["H"^(+)]. So:

$\left[{\text{H}}^{+}\right] = x = {10}^{- 4.60}$

Therefore:

$\textcolor{b l u e}{{K}_{a}} \approx {\left({10}^{- 4.60}\right)}^{2} / \left(0.1 - {10}^{- 4.60}\right) = \textcolor{b l u e}{6.310 \times {10}^{- 9}}$

Mar 31, 2017

First calculate Ka then we can calculate Kh

I'm going to find the Ka from a different way you may dont know which is only applicable to weak acids

When we solve for Ka

we do this

$\sqrt{\text{Ka" xx "molarity}} = m$

$- \log \left(m\right) = p H$

so by solving for m when given the pH we can find the answer of square root.

so -log(x) = 4.60

x = ${10}^{-} 4.60$
x = 0.00003

$\sqrt{\text{Ka" xx " 0.1M}} = 0.00002511886$

solve for Ka

$K a \times 0.1 = {0.00002511886}^{2}$

$6.30957344e-9 = K a$

${K}_{h} = {K}_{w} / \left(K b\right)$
and u will again get the Ka

For acids like this Ka = Kh

Oswald's law of dillution can be used to calculate Ka of weak electrolytes or weak acids like this one