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
$\therefore{\left\lbrack Al_{2}\left( SO_{4} \right)_{3} \right\rbrack\frac{4.8 \times 10^{- 3}}{3}^{- 3}M}_{\min}$
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