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Hunter-Best [27]
2 years ago
10

A 0.100 M solution of K2SO4 would contain the same total ion concentration as which of the following solutions?0.0800 M Na2CO3 0

.100 M NaCl 0.0750 M Na3PO4 0.0500 M NaOH
Chemistry
1 answer:
AysviL [449]2 years ago
7 0

Answer:

  • <em><u>The third chice: 0.0750 M Na₂SO₄</u></em>

Explanation:

Assume 100% ionization:

<em><u>1) 0.100 M solution K₂SO₄</u></em>

  • K₂SO4 (aq) → 2K⁺ (aq) + SO₄²⁻ (aq)

  • Mole ratios: 1 mol K₂SO4 : 2 mol K⁺  + 1 mol SO₄²⁻ (aq) : 3 mol ions. This is 1 : 3

  • At constant volume, the mole ratios are equal to the concentration ratios (M).

  • 1  M K₂SO₄: 3 M ions = 0.100 M K₂SO₄ / x ⇒ x = 0.300 M ions

This means, that you have to find which of the choices is a solution that contains the same 0.300 M ion concentration.

<u>2) 0.0800 M Na₂CO₃</u>

  • Na₂CO₃ (aq) → 2 Na⁺ + CO₃⁻

  • 1 M Na₂CO₃ / 3 M ions = 0.0800M / x ⇒ x = 0.0267 M ions

This is not equal to 0.300 M, so this solution would not contain the same total concentration as a 0.100 M solution of K₂SO₄, and is not the right answer.

<u>3)  0.100 M NaCl </u>

  • NaCl → Na⁺ + Cl⁻

  • 1 M NaCl / 2 M ions = 0.100 M NaCl / x ⇒ x = 0.200 M ions

This is not equal to 0.300 M ion, so not a correct option.

<u>4) 0.0750 M Na₃PO₄</u>

  • Na₃PO₄ → 3Na⁺ + PO₄³⁻

  • 1 M Na₃PO₄ / 4 M ions = 0.0750 M Na₃PO₄ / x ⇒ x = 0.300 M ions

Hence, this ion concentration is equal to the ion concentration of a 0.100 M solution of K₂SO₄, and is the correct choice.

<u>5)  0.0500 M NaOH </u>

  • NaOH → Na⁺ + OH⁻

  • 1 M NaOH / 2 mol ions = 0.0500 M NaOH / x ⇒ x = 0.100 M ions

Not equal to 0.300 M, so wrong choice.

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Answer:

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Explanation:

To find the limiting reactant we need to calculate the number of moles of each one:

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I hope it helps you!

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