Dear Student, Mg(OH)2 ↔ Mg2+ + 2OH- s1 2s1Ksp=(s1)(2s1)2=4s13and, 4s13=8 9×10-13or, s1=(8 9×10-134)13=6 05×10-5Al(OH)3 ↔ Al3+ + 3OH- s2 3s2Ksp=(s2)(3s2)3=27s24and, 27s24=5×10-33or, s2=(5×10-3327)14=3 69×10-9s1s2=6 05×10-53 69×10-9=1 64×104Hence, correct answer is (B)

Question-
Question- e soiubiliry product Mg(OH)2 in water at 25 oc is 8.9 q 10 13 (mole dm-3)3 while that Of is 5 q 10-33 (mol dm-3)4. If SJ and S? are the solubilities of Mg(OH)2 and in water in mol dm-3 at 25 oc, what is the order of magnitude of the ratio, SI/S ?

Dear Student,

Mg (OH)_{2} \leftrightarrow {Mg}^{2 +} + 2 {OH}^{-} s_{1} 2 s_{1} K_{sp} = (s_{1}) (2 s_{1})^{2} = 4 {s_{1}}^{3} and, 4 {s_{1}}^{3} = 8.9 \times 10^{- 13} or, s_{1} = (\frac{8.9 \times 10^{- 13}}{4})^{\frac{1}{3}} = 6.05 \times 10^{- 5} Al (OH)_{3} \leftrightarrow {Al}^{3 +} + 3 {OH}^{-} s_{2} 3 s_{2} K_{sp} = (s_{2}) (3 s_{2})^{3} = 27 {s_{2}}^{4} and, 27 {s_{2}}^{4} = 5 \times 10^{- 33} or, s_{2} = (\frac{5 \times 10^{- 33}}{27})^{\frac{1}{4}} = 3.69 \times 10^{- 9} \frac{s_{1}}{s_{2}} = \frac{6.05 \times 10^{- 5}}{3.69 \times 10^{- 9}} = 1.64 \times 10^{4} Hence, correct answer is (B)

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Question- Question- e soiubiliry product Mg(OH)2 in water at 25 oc is 8.9 q 10 13 (mole dm-3)3 while that Of is 5 q 10-33 (mol dm-3)4. If SJ and S? are the solubilities of Mg(OH)2 and in water in mol dm-3 at 25 oc, what is the order of magnitude of the ratio, SI/S ?

Question-
Question- e soiubiliry product Mg(OH)2 in water at 25 oc is 8.9 q 10 13 (mole dm-3)3 while that Of is 5 q 10-33 (mol dm-3)4. If SJ and S? are the solubilities of Mg(OH)2 and in water in mol dm-3 at 25 oc, what is the order of magnitude of the ratio, SI/S ?