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Orlov [11]
3 years ago
15

Determine the molar solubility of pbso4 in pure water. ksp (pbso4) = 1.82 x 10-8.

Chemistry
1 answer:
Misha Larkins [42]3 years ago
5 0
Use the ICE table approach as solution:

           PbSO₄   --> Pb²⁺ + SO₄²⁻
I             -                 0          0
C           -                +s         +s
E           -                  s          s

Ksp = [Pb²⁺][SO₄²⁻]
1.82×10⁻⁸ = s²
Solving for s,
s = <em>1.35×10⁻⁴ M</em>
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Answer:

155.16 g.

Explanation:

  • Firstly, It is considered as a stichiometry problem.
  • From the balanced equation: 2NaOH + H₂SO₄ → 2Na₂SO₄ + 2H₂O
  • It is clear that the stichiometry shows that 2.0 moles of NaOH reacts with 1.0 mole of H₂SO₄ to give 2.0 moles of Na₂SO₄ and 2.0 moles of H₂O.
  • We must convert the grams of water (57.0 g) to moles <em>(n = mass/molar mass)</em>.
  • n = (57.0 g) / (18.0 g/mole) = 3.1666 moles.
  • Now, we can get the number of moles of H₂SO₄ that is needed to produce 3.1666 moles of water.
  • <em>Using cross multiplication:</em>
  • 1.0 mole of H₂SO₄ → 2.0 moles of H₂O, <em>from the stichiometry of the balanced equation</em>.
  • ??? moles of H₂SO₄ → 3.1666 moles of H₂O.
  • The number of moles of H₂SO₄ that will produce 3.1666 moles of H₂O <em>(57.0 g)</em> is (1.0 x 3.1666 / 2.0) = 1.5833 moles.
  • Finally, we should convert the number of moles of H₂SO₄ into grams <em>(n = mass/molar mass)</em>.
  • Molar mass of H₂SO₄ = 98.0 g/mole.
  • mass = n x molar mass = (1.5833 x 98.0) = 155.16 g.
6 0
3 years ago
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Explanation:

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3 0
2 years ago
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