Ionic EquilibriumHard
Question
A sample of hard water contains 0.005 mole of CaCl2 per litre. What is the minimum concentration of Al2(SO4)3 which must be exceeded for removing Ca2+ ions from this water sample? The solubility product of CaSO4 is 2.4 × 10–5.
Options
A.4.8 × 10−3 M
B.1.6 × 10−3 M
C.0.0144 M
D.2.4 × 10−3 M
Solution
CaSO4 (s) $\rightleftharpoons$Ca2+ + SO42−
To start ppt, Q > Ksp
or, 0.005 × [SO42−] > 2.4 × 10–5
∴ [SO42−] > 4.8 × 10–3 M
$\therefore{\left\lbrack Al_{2}\left( SO_{4} \right)_{3} \right\rbrack\frac{4.8 \times 10^{- 3}}{3}^{- 3}M}_{\min}$
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
The volume of the water needed to dissolve 1 g of BaSO4 (KSP = 1.1 × 10-10) at 25oC is:...The conjugate acid of NH2- is :...When pure water is saturated with CaCO3 and CaC2O4, the concentration of calcium ion in the solution under equilibrium i...The pH value of N/10 NaOH is :...pH of a solution made by mixing 50 mL of 0.2 M NH4Cl and 75 mL of 0.1 M NaOH is :-[pKb of NH3(aq) = 4.74. log 3 = 0.47]...