# Question d7354

Aug 22, 2017

$\text{pH = 3.40"; ["H"_2"A"] = "0.0646 mol/L"; ["A"^"2-"] = 8.7 × 10^"-9" color(white)(l)"mol/L}$

#### Explanation:

pH of solution

To a first approximation, we can assume that only the first ionization is important in determining the pH.

$\textcolor{w h i t e}{m m m m m m m} \text{H"_2"A" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HA"^"-}$
$\text{I/mol/L:} \textcolor{w h i t e}{m m} 0.0650 \textcolor{w h i t e}{m m m m m m l} 0 \textcolor{w h i t e}{m m m l} 0$
$\text{C/mol/L:"color(white)(mmm)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mmll)"+} x$
$\text{E/mol/L:"color(white)(ml)"0.0650-} x \textcolor{w h i t e}{m m m m m l l} x \textcolor{w h i t e}{m m m l} x$

K_text(a1) = (["H"_3"O"^"+"]["HA"^"-"])/(["H"_2"A"]) = x^2/("0.0650-"x) = 2.4 × 10^"-6"

0.0650/(2.4 × 10^"-6") = 2.7 × 10^4 ≫ 400. ∴ x ≪0.0650

Then

x^2/0.0650 = 2.4 × 10^"-6"

x^2 = 0.0650 × 2.4 × 10^"-6" = 1.56 × 10^"-7"

x = 3.95 × 10^"-4"

["H"_3"O"^"+"] = ["HA"^"-"] = 3.95 × 10^"-4" color(white)(l)"mol/L"

"pH" = -log["H"_3"O"^"+"] = -log(3.95 × 10^"-4") = 3.40

Equilibrium concentration of $\text{H"_2"A}$

["H"_2"A"] = "(0.0650-"x") mol/L" = (0.0650 - 3.95×10^"-4"") mol/L" = "0.0646 mol/L"

Equilibrium concentration of $\text{A"^"2-}$

For this, we can use the second ionization step.

$\text{HA"^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"2-}$

${K}_{\textrm{a 2}} = \left(\left[\text{H"_3"O"^"+"]["A"^"2-"])/(["HA"^"-}\right]\right)$

["A"^"2-"] = (K_text(a2)["HA"^"-"])/(["H"_3"O"^"+"]) = (8.7 × 10^"-9" × color(red)(cancel(color(black)(3.95 × 10^"-4"))))/(color(red)(cancel(color(black)(3.95 × 10^"-4")))) = 8.7 × 10^"-9"color(white)(l) "mol/L"#