# What is the Ka of an acid of 6M which has a pH of 2.5 at 277.5K ?

Mar 17, 2017

K_text(a) = 2 × 10^"-6"

#### Explanation:

I would argue that the temperature has nothing to do with the value of ${K}_{\textrm{a}}$.

In this problem, it is the $\text{pH}$ and the initial concentration of the acid that determine the value of ${K}_{\textrm{a}}$.

Let's set up an ICE table to solve this problem.

$\textcolor{w h i t e}{m m m m m m m} \text{HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m l} 6 \textcolor{w h i t e}{m m m m m m l} 0 \textcolor{w h i t e}{m m m} 0$
$\text{C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmm)"+"xcolor(white)(mml)"+} x$
$\text{E/mol·L"^"-1":color(white)(ml)"6 -} \textcolor{w h i t e}{l} x \textcolor{w h i t e}{m m m m m l} x \textcolor{w h i t e}{m m l l} x$

We must use the $\text{pH}$ to calculate the value of $x$.

$\text{pH = 2.5}$

["H"_3"O"^"+"] = 10^"-2.5"color(white)(l) "mol/L" = "0.0031 mol/L" = x

The ${K}_{\text{a}}$ expression is:

${K}_{\text{a" = (["A"^"-"]["H"_3"O"^"+"])/(["HA"]) = (x × x)/(0.100-x) = x^2/(0.100-x) = 0.0031^2/(6 - 0.0031) = (1.00 × 10^"-5")/6 = 2 × 10^"-6}}$