Chemistry, asked by Nobita142, 9 months ago

saturated cac2o4 (aq) is prepared and 250 ml of the solution is withdrawn andtitrated with 6.3 ml of 0.00102 m kmno4 (aq) what is the value of ksp for cac2o4?

Answers

Answered by qwsuccess
1

Given:

  • Volume of saturated solution of CaC₂O₄ (V) = 0.25 L
  • Volume of KMnO₄ used in titration (V1) = 6.3 mL
  • Molarity of KMnO₄ used (M) = 0.00102 M

To find:

The Ksp of CaC₂O₄.

Solution:

  • The amount of oxalate ion that will react with KMnO₄ will be equal to the amount of  oxalate ions soluble (S),
  • Moles of KMnO₄ reacted = M*V1
  • One mole of KMnO₄reacts with 2.5 moles of oxalate ions. So, moles of soluble oxalate ions (n) = 2.5*M*V1 = 1.60*10⁻⁵
  • S = n/V = 6.40*10⁻⁵ M
  • Ksp = [Ca²⁺][C₂O₄²⁻] = S² = 4.096*10⁻⁹

Answer:

The Ksp of CaC₂O₄ = 4.096*10⁻⁹

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