# What is the pH of 0.05 mol/L acetic acid? What are the hydronium ion and acetate ion concentrations in a 0.05 mol/L pH 4 acetate buffer?

Mar 19, 2017

(a) $\text{pH = 4.77}$; (b) ["H"_3"O"^"+"] = 1.00 × 10^"-4" color(white)(l)"mol/dm"^3; (c) ["A"^"-"] = "0.16 mol·dm"^"-3"

#### Explanation:

(a) pH of aspirin solution

Let's write the chemical equation as

$\textcolor{w h i t e}{m m m m m m m m l} \text{HA" color(white)(m)+color(white)(m) "H"_2"O" ⇌ "H"_3"O"^"+"color(white)(m) + color(white)(ml)"A"^"-}$
$\text{I/mol·dm"^"-3} : \textcolor{w h i t e}{m m} 0.05 \textcolor{w h i t e}{m m m m m m m m l} 0 \textcolor{w h i t e}{m m m m m l l} 0$
$\text{C/mol·dm"^"-3":color(white)(mml)"-"xcolor(white)(mmmmmmmm)"+"xcolor(white)(mlmmml)"+} x$
$\text{E/mol·dm"^"-3":color(white)(m) "0.05 -} \textcolor{w h i t e}{l} x \textcolor{w h i t e}{m m m m m m m l} x \textcolor{w h i t e}{m m x m m m} x$

K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = x^2/("0.05 -"color(white)(l)x) = 3.27 × 10^"-4"

Check for negligibility

0.05/(3.27 × 10^"-4") = 153 < 400

$x$ is not less than 5 % of the initial concentration of $\left[\text{HA}\right]$.

We cannot ignore it in comparison with 0.05, so we must solve a quadratic.

Then

x^2/(0.05 -x) = 3.27 × 10^"-4"

x^2 = 3.27×10^"-4"(0.05-x)= 1.635 × 10^"-5" - 3.27 × 10^"-4"x

x^2 + 3.27 × 10^"-4"x -1.635 ×10^"-5" =0

x = 1.68 × 10^"-5"

["H"_3"O"^"+"] = x color(white)(l)"mol/L" = 1.68 × 10^"-5"color(white)(l) "mol/L"

"pH" = "-log"["H"_3"O"^"+"] = "-log"(1.68 × 10^"-5") = 4.77

(b) $\left[\text{H"_3"O"^"+}\right]$ at pH 4

["H"_3"O"^"+"] = 10^"-pH" color(white)(l)"mol/L" = 1.00 ×10^"-4"color(white)(l) "mol/L"

(c) Concentration of $\text{A"^"-}$ in the buffer

We can now use the Henderson-Hasselbalch equation to calculate the $\left[\text{A"^"-}\right]$.

"pH" = "p"K_"a" + log((["A"^"-"])/(["HA"]"))

4.00 = -log(3.27 × 10^"-4") + log((["A"^"-"])/0.05) = 3.49 + log((["A"^"-"])/0.05)

log((["A"^"-"])/0.05) = "4.00 - 3.49" = 0.51

$\frac{\left[\text{A"^"-}\right]}{0.05} = {10}^{0.51} = 3.24$

["A"^"-"] = 0.05 × 3.24 = 0.16

The concentration of $\text{A"^"-}$ in the buffer is 0.16 mol/L.