saturated cac2o4 (aq) is prepared and 250 ml of the solution is withdrawn andtitrated with 6.3 ml of 0.00102 m kmno4 (aq) what is the value of ksp for cac2o4?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