# 50.0 mL of 0.100 M "NH"_3 is titrated with 0.100 M "HCl". What is the pH of the "NH"_3/"HCl" mixture after the addition of 25.0 mL of "HCl"?

## The given ${\text{K}}_{b}$ for ${\text{NH}}_{3}$ to use, is $1.8 \times {10}^{-} 5$

Jul 15, 2018

After reading this, take the time to compare this answer with this one. It is very important that you see these are the same situation with the roles reversed...

Spotting patterns like this will help. We again find that

${\text{pH" = "pK}}_{a}$

at the half-equivalence point, i.e.

$\text{pH} = 9.26$

And this situation is the formation of a buffer.

BEFORE DOING THE PROBLEM

There is less strong acid than weak base (instead of less strong base than weak acid). Hence, before even starting the problem, we should recognize that there is more ${\text{NH}}_{3}$ than $\text{HCl}$.

As a result, a buffer must form because ${\text{NH}}_{3}$ will remain and ${\text{NH}}_{4}^{+}$ (its conjugate acid) will form by the reaction of $\text{HCl}$ with NOT all of the ${\text{NH}}_{3}$.

${\text{NH"_3(aq) + "HCl"(aq) -> "NH"_4^(+)(aq) + "Cl}}^{-} \left(a q\right)$

Before doing any major math, we can predict that since the volume of $\text{HCl}$ is half the volume of ${\text{NH}}_{3}$, half the ${\text{NH}}_{3}$ is neutralized and we get to the half-equivalence point, where $\left[{\text{NH"_4^(+)] = ["NH}}_{3}\right]$.

Hence, if we do not get ${\text{pH" = "pK}}_{a}$ and the above equivalence, we missed something. Now we can actually try the math.

INITIAL CONCENTRATIONS

Weak base:

(0.0500 cancel"L" xx "0.100 mol"/cancel"L")/("0.0500 L" + "0.0250 L") = "0.00500 mols NH"_3/"0.0750 L"

$= {\text{0.066 M NH}}_{3}$

Strong acid:

(0.0250 cancel"L" xx "0.100 mol"/cancel"L")/("0.0500 L" + "0.0250 L") = "0.00250 mols HCl"/"0.0750 L"

$= {\text{0.033 M HCl" = "0.033 M H}}^{+}$

And now these react, strong acid with weak base, just like $\text{NaOH}$ (strong base) with $\text{HCN}$ (weak acid).

${\text{0.00500 mols NH"_3 - "0.00250 mols HCl" = "0.00250 mols NH}}_{3}$ used up,

$\implies {\text{0.00250 mols NH"_3/"0.0750 L" = "0.033 M NH}}_{3}$ remains.

and therefore, ${\text{0.00250 mols NH}}_{4}^{+}$ forms by the $\text{HCl}$ protonating the ${\text{NH}}_{3}$ ($\text{HCl}$ acting as a Bronsted-Lowry acid to the Lewis base ${\text{NH}}_{3}$).

$\implies {\text{0.00250 mols NH"_4^(+)/"0.0750 L" = "0.033 M NH}}_{4}^{+}$ forms.

Of course, we expected this and confirmed it: $\left[N {H}_{4}^{+}\right] = \left[N {H}_{3}\right]$ and we are at the half-equivalence point.

FINDING pH

Now, we could approach this using either ${K}_{b}$ or ${K}_{a}$, but we must write the correct reaction in water whichever way we do it. By now you've seen the ICE table several times, so I'll let you fill it in this time.

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

We can proceed to the mass action expression:

${K}_{b} = 1.8 \times {10}^{- 5} = \left(\left[{\text{NH"_4^(+)]["OH"^(-)])/(["NH}}_{3}\right]\right)$

$= \frac{\left(0.033 + x\right) x}{0.033 - x}$

Small x approximation:

$1.8 \times {10}^{- 5} \approx \frac{\left(0.033 + \cancel{x}\right) x}{0.033 - \cancel{x}}$

$\implies x \approx {K}_{b} = 1.8 \times {10}^{- 5} \text{M}$

Since $x \equiv \left[{\text{OH}}^{-}\right]$,

$- \log x = - \log \left({K}_{b}\right)$

=> ul"pOH" = -log["OH"^(-)] = ul( "pK"_b)

Therefore, since

${\text{pK"_a + "pK"_b = "pK}}_{w}$

${\text{pH" + "pOH" = "pK}}_{w}$

we find that at any temperature, if $\left[{\text{HA"] = ["A}}^{-}\right]$,

$\textcolor{b l u e}{{\text{pH") = "pK"_w - "pOH" = "pK"_w - "pK"_b = color(blue)("pK}}_{a}}$

You can do it from here.