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?

Question

Natalie Anton · Accepted Answer

The solution that holds the greatest concentration of ions is A. #"K"_2"SO"_4#.

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")# There are 3.00 mol/L of ions present. B. #"H"_2"SO"_4# #"K"_2"SO"_4# ionizes in two steps. While the second step is not completed, the first one is. #"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. Let's assume for the purposes of argument that the ion concentration 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. Let's assume for the purposes of argument that the ion concentration 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"# Compared to phosphoric acid, it is weaker. The concentration of ions is 0.05 mol/L because 1 mol of acetic acid may contribute 0.05 mol of ions.

Jeremiah Adams · Answer

undefined The solution containing the greatest concentration of ions would be the one with the highest number of ions per mole of solute. To determine this, we need to calculate the number of ions produced by each compound when dissolved in water.

1. NaCl (sodium chloride) dissociates into 2 ions (Na⁺ and Cl⁻) per formula unit.
   Number of ions = 2 ions/mol

2. MgCl₂ (magnesium chloride) dissociates into 3 ions (Mg²⁺ and 2Cl⁻) per formula unit.
   Number of ions = 3 ions/mol

3. AlCl₃ (aluminum chloride) dissociates into 4 ions (Al³⁺ and 3Cl⁻) per formula unit.
   Number of ions = 4 ions/mol

Therefore, AlCl₃ would produce the greatest concentration of ions when 1.00 mol of each compound is dissolved in enough water to yield a 10.0 L solution.