# Calculate the pH of a "0.2 M" solution of "NaC"_2"H"_3"O"_2 given that the hydrolysis constant is K_h = 5.6xx10^(-10)?

May 24, 2018

pH = 9.02

#### Explanation:

We can use an ICE table to help with the calculation.

$\textcolor{w h i t e}{m m m m m m m} \text{A"^"-" + "H"_2"O" ⇌ "HA" + "OH"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m l l} 0.2 \textcolor{w h i t e}{m m m m m m} 0 \textcolor{w h i t e}{m m l l} 0$
$\text{C/mol·L"^"-1":color(white)(mll)"-"xcolor(white)(mmmmml)"+"xcolor(white)(mm)"+} x$
$\text{E/mol·L"^"-1":color(white)(m)"0.2-} x \textcolor{w h i t e}{m m m m m l} x \textcolor{w h i t e}{m m l l} x$

K_text(h) = (["HA"]["OH"^"-"])/(["A"^"-"]) = 5.6 × 10^"-10"

x^2/("0.2-"x) = 5.6 × 10^"-10"

Apply the 5 % rule

0.2/(5.6 × 10^"-10") = 3.6 × 10^8 > 400.

x ≪0.2

Then

x^2/0.2 = 5.6 × 10^"-10"

x^2 = 0.2(5.6 × 10^"-10") = 1.1 × 10^"-10"

x = 1.1 × 10^"-5"

["OH"^"-"] = x color(white)(l)"mol/L" = 1.1 × 10^"-5"color(white)(l) "mol/L"

"pOH" = "-log"["OH"^"-"] = "-log"(1.1 × 10^"-5") = 4.98

$\text{pH = 14.00 - pOH = 14.00 - 4.98 = 9.02}$