# Question 721ab

Mar 14, 2017

pH = 1.81

#### Explanation:

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

$\textcolor{w h i t e}{m m m m m m m} \text{HF"color(white)(l) +color(white)(l) "H"_2"O"color(white)(l) ⇌ color(white)(l) "H"_3"O"^+color(white)(l) +color(white)(l) "F"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m l l} 0.35 \textcolor{w h i t e}{m m m m m m m l l} 0 \textcolor{w h i t e}{m m m l l} 0$
$\text{C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmmml)"+x"color(white)(mmm)"+x}$
$\text{E/mol·L"^"-1":color(white)(ll)"0.35 -} x \textcolor{w h i t e}{m m m m m m m} x \textcolor{w h i t e}{m m m l l} x$

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

${K}_{\text{a" = (["H"_3"O"^+]["F"^"-"])/(["HF"]) = 6.8 × 10^"-4}}$

K_a = (x × x)/(0.35-x) = x^2/(0.35-x) = 6.8 × 10^"-4"

0.35/K_"a" = 0.35/(6.31 × 10^"-4") = 515 > 400

x ≪ 0.35, and the equation becomes

x^2/0.35 = 6.8 × 10^"-4"

x^2 = 0.35 × 6.8 × 10^"-4" = 2.38 × 10^"-4"

x = 1.54 × 10^"-2"

["H"^+] = 1.54 × 10^"-2"color(white)(l) "mol/L"

"pH" = "-log"["H"_3"O"^"+"] = "-log"(1.54 × 10^"-2") = 1.81#