What is the pH of the solution formed by mixing 100 mL of 0.5 mol/L ammonia with 50 mL of 1 M hydrochloric acid?

For ammonia. $\text{p} {K}_{\textrm{a}} = 9.25$.

Nov 27, 2017

pH = 4.9

Explanation:

Step 1. Figure out what is happening

The $\text{p} {K}_{\textrm{a}}$ values tell you that $\text{HCl}$ is a strong acid, that $\text{NH"_4^"+}$ is a weak acid, and that ${\text{NH}}_{3}$ is a weak base.

Thus, when you mix the strong acid $\text{HCl}$ with the weak base ${\text{NH}}_{3}$, the reaction will go to completion.

${\text{Initial moles of NH"_3 = 0.100 color(red)(cancel(color(black)("L NH"_3))) × "0.5 mol NH"_3/(1 color(red)(cancel(color(black)("L NH"_3)))) = "0.050 mol NH}}_{3}$

$\text{Initial moles of HCl" = 0.050 color(red)(cancel(color(black)("L HCl"))) × "1 mol HCl"/(1 color(red)(cancel(color(black)("L HCl")))) = "0.050 mol HCl}$

$\textcolor{w h i t e}{m m m m m l} \text{NH"_3 +color(white)(l) "HCl" → "NH"_4"Cl}$
$\text{I/mol} : \textcolor{w h i t e}{m l l} 0.050 \textcolor{w h i t e}{m l} 0.050 \textcolor{w h i t e}{m m m l} 0$
$\text{C/mol":color(white)(m)"-0.050"color(white)(m)"-0.050"color(white)(m)} + 0.050$
$\text{E/mol} : \textcolor{w h i t e}{m m l l} 0 \textcolor{w h i t e}{m m m l} 0 \textcolor{w h i t e}{m m m l} 0.050$

The ammonia and hydrochloric acid react completely.

You now have 0.050 mol of $\text{NH"_4"Cl}$ in 150 mL of solution.

Step 2. Calculate the pH of the solution

$\text{Initial concentration of NH"_4^"+" = "0.050 mol"/"0.150 L" = "0.33 mol/L}$

Now, we can use an ICE table.

$\textcolor{w h i t e}{m m m m m m l} \text{NH"_4^"+" + "H"_2"O" ⇌ "NH"_3 + "H"_3"O"^"+}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m l} 0.33 \textcolor{w h i t e}{m m m m m m l} 0 \textcolor{w h i t e}{m m m l} 0$
$\text{C/mol·L"^"-1":color(white)(mll)"-"xcolor(white)(mmmmmm)"+"xcolor(white)(mml)"+} x$
$\text{E/mol·L"^"-1":color(white)(m)"0.33-} x \textcolor{w h i t e}{m m m m m l} x \textcolor{w h i t e}{m m m l} x$

${K}_{\textrm{a}} = \left(\left[\text{NH"_3]["H"_3"O"^"+"])/(["NH"_4^"+}\right]\right) = {x}^{2} / \left(0.33 - x\right)$

$\text{p} {K}_{\textrm{a}} = 9.25$

K_text(a) = 10^("-p"K_text(a)) = 10^"-9.25" = 5.62 × 10^"-10"

x^2/(0.33-x) = 5.62 × 10^"-10"

0.33/(5.62 × 10^"-10") = 5.9 ×10^8, so x≪ 0.33

Then,

x^2/0.33 = 5.62 × 10^"-10"

x^2 = 0.33 × 5.62 × 10^"-10" = 1.9 × 10^"-10"

x = 1.4 × 10^"-5"

["H"_3"O"^"+"] = xcolor(white)(l) "mol/L" = 1.4 × 10^"-5"color(white)(l) "mol/L"

"pH" = "-log"["H"_3"O"^"+"] = -log(1.4 × 10^"-5") =4.9