A. #"K"_2"SO"_4#
#"K"_2"SO"_4# is a strong electrolyte. It ionizes completely in solution.
#underbrace("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. #"H"_2"SO"_4#
#"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#
#"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 #"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. #"H"_3"PO"_4#
#"H"_3"PO"_4# is a weak acid. It ionizes in three steps.
#"H"_3"PO"_4 + "H"_2"O" ⇌ "H"_3"O"^"+" + "H"_2"PO"_4^"-"; K_1 = 7.1 × 10^"-3"#
#"H"_2"PO"_4^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HPO"_4^"2-"; color(white)(l)K_2 = 6.3 × 10^"-8"#
#"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 #"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. #"CH"_3"COOH"#
#"CH"_3"COOH"# is a weak acid.
#"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.
