When 1.00 mol of each compound below is dissolved in enough water to yield a 10.0 L solution, which solution contains the greatest concentration of ions?

Jun 25, 2017

The solution that holds the greatest concentration of ions is A. ${\text{K"_2"SO}}_{4}$.

Explanation:

A. ${\text{K"_2"SO}}_{4}$

${\text{K"_2"SO}}_{4}$ is a strong electrolyte. It ionizes completely in solution.

$\underbrace{\text{K"_2"SO"_4)_color(red)("1.00 mol/L") → underbrace("2K"^"+" + "SO"_4^"2-")_color(red)("3.00 mol/L}}$

The concentration of ions is 3.00 mol/L.

B. ${\text{H"_2"SO}}_{4}$

${\text{K"_2"SO}}_{4}$ ionizes in two steps.

The first step goes to completion, but the second step does not.

"H"_2"SO"_4 + "H"_2"O" → "H"_3"O"^"+" + "HSO"_4^"-"; K_1 = 3.0 ×10^3
$\text{HSO"_4^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "SO"_4^"2-"; color(white)(m)K_2 = 1.0 × 10^"-2}$

Thus, a 1.00 mol/L solution of ${\text{H"_2"SO}}_{4}$ contributes 2 mol of ions from the first step plus some more from the second ionization.

For argument's sake, let's say that the concentration of ions is 2.1 mol/L.

C. ${\text{H"_3"PO}}_{4}$

${\text{H"_3"PO}}_{4}$ is a weak acid. It ionizes in three steps.

$\text{H"_3"PO"_4 + "H"_2"O" ⇌ "H"_3"O"^"+" + "H"_2"PO"_4^"-"; K_1 = 7.1 × 10^"-3}$
$\text{H"_2"PO"_4^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HPO"_4^"2-"; color(white)(l)K_2 = 6.3 × 10^"-8}$
$\text{HPO"_4^"2-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "PO"_4^"3-"; color(white)(ml)K_3 = 4.2 × 10^"-13}$

Thus, a 1.00 mol/L solution of ${\text{H"_3"PO}}_{4}$ contributes perhaps 0.5 mol of ions.

For argument's sake, let's say that the concentration of ions is 0.5 mol/L.

D. $\text{CH"_3"COOH}$

$\text{CH"_3"COOH}$ is a weak acid.

$\text{CH"_3"COOH" + "H"_2"O" ⇌ "H"_3"O"^"+" + "CH"_3"COO"^"-"; K_text(a) = 1.76 × 10^"-5}$

It is weaker than phosphoric acid.

1 mol of acetic acid might contribute 0.05 mol of ions, so the concentration of ions is 0.05 mol/L.