Question

When 15.0 mL of a 8.39x10 * M calcium acetate solution is combined with 22.0 mL of a 3.88*10-4 M potassium sulfate solution does a precipitate form? (yes or no) For these conditions the Reaction Quotient, Q, is equal to

Answer 1

Ans :-

No. of moles of Ca(CH3COO)2 = Molarity x Volume in L

= 8.39 x 10-4 M x 0.015 L

= 1.2585 x 10-5 mol

No. of moles of K2SO4 = Molarity x Volume in L

= 3.88 x 10-4 M x 0.022 L

= 8.536 x 10-6 mol

Total volume after mixing = 0.015 L+ 0.022 L = 0.037 L

Molarity of Ca(CH3COO)2 after mixing = Moles/Volume in L

= 1.2585 x 10-5 mol / 0.037 L

= 3.40 x 10-4 M

Also,

Molarity of K2SO4 after mixing = Moles/Volume in L

= 8.536 x 10-6 mol / 0.037 L

= 2.31 x 10-4 M

Partial dissociation of CaSO4 in aqueous solution is :

CaSO4 (s) <-----------------------> Ca2+ (aq) + SO42- (aq)

Reaction quotient (Q) (Which is equal to the product of the molar concentration of products raised to the power of stoichiometric coefficient at any stage of the reaction). Similarly, Solubility product (Ksp) (Which is equal to the product of the molar concentration of products raised to the power of stoichiometric coefficient at equilibrium stage of the reaction). Ksp value of CaSO4 = 2.4 x 10-5 M2 If Q < Ksp, then precipitates are not form. If Q > Ksp, then precipitates are form. If Q = Ksp​​​​​​​, then reaction is in equilibrium stage.

Expression of reaction quotient (Q) is :

Q = [Ca2+].[SO42-]

= (3.40 x 10-4 M).(2.31 x 10-4 M)

= 7.85 x 10-8 M2

Because, Q < Ksp, therefore precipitates are not formed. (NO)