# HA is a weak acid. A 0.21M solution of HA has a pH of 2.68. What is the Ka value for HA?

Jun 18, 2018

 K_text(a) = 2.1 × 10^"-5"

#### Explanation:

$\text{pH = 2.68}$

["H"_3"O"^"+"] = 10^"-pH" color(white)(l)"mol/L" = 10^"-2.68"color(white)(l)"mol/L" = "0.002 09 mol/L"

We can use an ICE table to help us calculate the value of ${K}_{\textrm{a}}$.

$\textcolor{w h i t e}{m m m m m m l l} \text{HA" + "H"_2"O" ⇌ "H"_3"O"^"+" + color(white)(l)"A"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m l l} 0.21 \textcolor{w h i t e}{m m m m m m} 0 \textcolor{w h i t e}{m m m l} 0$
$\text{C/mol·L"^"-1": color(white)(ml)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mml)"+} x$
$\text{E/mol·L"^"-1": color(white)(m)"0.21-} x \textcolor{w h i t e}{m m m m m l} x \textcolor{w h i t e}{m m m l} x$

At equilibrium,

["H"_3"O"^"+"] = "0.002 09 mol/L"

$x = \text{0.002 09}$

K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = x^2/(0.21-x) = "0.002 09"^2/("0.21-0.002 09") = (4.16 × 10^"-6")/0.208

= 2.1 × 10^"-5"