You are provided with a 500 mL of hard water, containing 0.005 mole of CaCl2, and two sulphuric acid samples of 0.001 M and 0.02 M concentrations. The solubility product of calcium sulphate in water at 25∘C is 2.4 × 10−5. Choose the most appropriate answer:
You are provided with a 500 mL of hard water, containing 0.005 mole of CaCl2, and two sulphuric acid samples of 0.001 M and 0.02 M concentrations. The solubility product of calcium sulphate in water at 25∘C is 2.4 × 10−5. Choose the most appropriate answer:
Both samples of 
will cause precipitation
will cause precipitation but 
sample doesn't
will cause precipitation but 
sample doesn't
The correct option is B will cause precipitation but sample doesn't
There is a major difference between the previous problem and this one! Here, there is no mention of the volume of the acid solutions at all. Without the volume it is impossible to proceed. Let us assume for the sake of simplicity that the we mix 500 mL of the acid in every case.
Concentration of [Ca2+] before= 0.01 M (why?)
Let us first evaluate the ionic product of CaSO4 for 0.001 M acid solution
Concentration of [Ca2+] after mixing= 0.005 M (why?)
After mixing, the volume doubles while the number of moles is the same as before! Similarly, the [SO2−4] = 0.0012 = 0.0005 M
Ionic product of Ca2+ and SO2−4 = [Ca2+][SO42-] = 0.005 × 0.0005 = 2.5 × 10−6 < Ksp (CaSO4) = 2.4 × 10−5 Hence the 0.001 M Sulphuric acid will not cause precipitation. So we can rule out two options already.
0.02 M sulphuric acid:
[Ca2+] in solution after adding H2SO4 = 0.0102 = 0.005 M
[SO2−4] in 0.02 M H2SO4 = 0.02 M
[SO2−4] in 1 L solution = 0.022 = 0.01 M
Ionic product = [Ca2+][SO2−4] = 0.005 × 0.01 = 5 × 10−5 > Ksp (CaSO4) = 2.4 × 10−5
So 0.02 M H2SO4 sample will cause precipitation.
will cause precipitation but sample doesn't
There is a major difference between the previous problem and this one! Here, there is no mention of the volume of the acid solutions at all. Without the volume it is impossible to proceed. Let us assume for the sake of simplicity that the we mix 500 mL of the acid in every case.
Concentration of [Ca2+] before= 0.01 M (why?)
Let us first evaluate the ionic product of CaSO4 for 0.001 M acid solution
Concentration of [Ca2+] after mixing= 0.005 M (why?)
After mixing, the volume doubles while the number of moles is the same as before! Similarly, the [SO2−4] = 0.0012 = 0.0005 M
Ionic product of Ca2+ and SO2−4 = [Ca2+][SO42-] = 0.005 × 0.0005 = 2.5 × 10−6 < Ksp (CaSO4) = 2.4 × 10−5 Hence the 0.001 M Sulphuric acid will not cause precipitation. So we can rule out two options already.
0.02 M sulphuric acid:
[Ca2+] in solution after adding H2SO4 = 0.0102 = 0.005 M
[SO2−4] in 0.02 M H2SO4 = 0.02 M
[SO2−4] in 1 L solution = 0.022 = 0.01 M
Ionic product = [Ca2+][SO2−4] = 0.005 × 0.01 = 5 × 10−5 > Ksp (CaSO4) = 2.4 × 10−5
So 0.02 M H2SO4 sample will cause precipitation.
