Question
Question: What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? ( For cal...
What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? ( For calcium sulphate, Ksp= 9.1×10−6 ).
Solution
The solubility product of an electrolyte is equal to the product of ionic concentrations of the electrolyte and for calcium sulphate the solubility product is as follows.
Ksp=[Ca2+][SO42−]
Ksp = Solubility product
[Ca2+] = Concentration of the calcium
[SO42−] = Concentration of the sulphate
Complete step by step solution:
- In the question it is asked to find the minimum volume of water required to dissolve 1 g of calcium sulphate.
- The given solubility product of calcium sulphate is Ksp= 9.1×10−6
- The given mass of the calcium sulphate is 1 g.
- When calcium is added to water it dissolves and the ionic parts are in equilibrium with the calcium sulphate and it is represented as follows.
CaSO4⇄Ca2++SO42−
- The solubility product of the above chemical equation is as follows.
Ksp=[Ca2+][SO42−]
Ksp = Solubility product
[Ca2+] = Concentration of the calcium
[SO42−] = Concentration of the sulphate
- Assume the solubility of the calcium sulphate is S
- Then