Q.17 The solubility product (Ksp) of Ca3(PO4)2 is 1.3 × 10−32. In a 0.02 M solution of Ca(NO3)2, the solubility of Ca3(PO4)2 (in units of M) is
(A) 6.5 × 10−31
(B) 1.6 × 10−26
(C) 8.0 × 10−16
(D) 4.0 × 10−14
Solubility of Ca3(PO4)2 in 0.02 M Ca(NO3)2
Problems based on solubility product (Ksp) and the
common ion effect are frequently asked in competitive examinations.
In this question, we calculate the
solubility of Ca3(PO4)2
in a 0.02 M Ca(NO3)2 solution.
Given
Ksp of Ca3(PO4)2 =1.3 × 10−32
Concentration of Ca(NO3)2 =0.02 M
Find the solubility of Ca3(PO4)2 (in M).
Step 1: Dissociation of Ca3(PO4)2
Ca3(PO4)2 (s)⇌ 3Ca2+ + 2PO43−
Step 2: Ksp Expression
Ksp =[Ca2+]3[PO43−]2
Step 3: Apply Common Ion Effect
Ca(NO3)2 is a strong electrolyte and provides
Ca2+ ions.
Initial [Ca2+] = 0.02 M
Let solubility of Ca3(PO4)2 = s
- [PO43−] = 2s
- [Ca2+] ≈ 0.02 (since 3s ≪ 0.02)
Step 4: Substitute Values
1.3 × 10−32 =(0.02)3 × (2s)2
1.3 × 10−32 =3.2 × 10−5 s2
Step 5: Calculate Solubility
s2 =1.3 × 10−32 /3.2 × 10−5
s2 = 4.06 × 10−28
s ≈ 2.0 × 10−14 M
Closest option = 4.0 × 10−14 M
Final Answer
Correct Option: (D) 4.0 × 10−14 M
Conclusion
The solubility of Ca3(PO4)2 decreases significantly
in the presence of Ca2+ ions due to the common ion effect.
Correct application of Ksp leads to the accurate answer.
