# How do you find the pH?

## pKa = 10.20 Concentration of Acidic Solution = 0.8470

Apr 2, 2016

$\text{pH = 5.14}$

#### Explanation:

We know that $\text{p"K_"a} = 10.20$.

${K}_{\text{a" = 10^"-10.20" = 6.31 × 10^"-11}}$

Let's set up an ICE table for the calculation of $\left[{\text{H"_3"O}}^{+}\right]$.

$\textcolor{w h i t e}{m m m m m m m} \text{HA"color(white)(l) +color(white)(l) "H"_2"O"color(white)(l) ⇌ color(white)(l) "H"_3"O"^+color(white)(l) +color(white)(l) "A"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m} 0.8470 \textcolor{w h i t e}{m m m m m m m} 0 \textcolor{w h i t e}{m m m m l} 0$
$\text{C/mol·L"^"-1":color(white)(mm)"-} x \textcolor{w h i t e}{m m m m m m l l} + x \textcolor{w h i t e}{m m l l} + x$
$\text{E/mol·L"^"-1":color(white)(ll)"0.8470 -} x \textcolor{w h i t e}{m m m m m l l} x \textcolor{w h i t e}{m m m m l} x$

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

${K}_{\text{a" = (["H"_3"O"^+]["A"^"-"])/(["HA"]) = 6.31 × 10^"-11}}$

K_a = (x × x)/(0.8470-x) = x^2/(0.8470-x) = 6.31 × 10^"-11"

Check if x ≪ 0.8470:

0.8470/K_"a" = 0.8470/(6.31 × 10^"-11") = 1.34 × 10^10 ≫ 400

x ≪ 0.8470, and the equation becomes

x^2/0.8470 = 6.31 × 10^"-11"

x^2 = 0.8470 × 6.31 × 10^"-11" = 5.34 × 10^"-11"

x = 7.31 × 10^"-6"

["H"^+] = 7.31 × 10^-6 "mol/L"

"pH" = -log["H"^+] = -log(7.31 × 10^-6) = 5.14