First slide

Ionic equilibrium

Question

Which of the following salts has the greatest molar solubility in pure water?

Very difficult

A

${CaCO}_{3} (K_{sp} = 8 .7 \times 10^{- 9})$
B

$CuS (K_{sp} = 8 .5 \times 10^{- 45})$
C

${Ag}_{2} {CO}_{3} (K_{sp} = 6 .2 \times 10^{- 12})$
D

${Pb}_{9} {IO}_{3} (K_{sp} = 2 .6 \times 10^{- 13})$

Solution

For CaCO₃, solubiliry can be calculated as
${CaCO}_{3} ⇌ {Ca}_{S}^{2 +} + \underset{S}{{CO}_{3}^{2}} K_{sp} = S \cdot S = S^{2} \Rightarrow S = \sqrt{K_{sp}} = \sqrt{8 .7 \times 10^{- 9}} = 0 .0051 = 9 .3 \times 10^{- 5}$
For CuS, solubility can be calculated as
$CuS ⇌ {Cu}^{2 +} + S_{S}^{2 -} K_{sp} = S \cdot S = S^{2} S = \sqrt{K_{sp}} = \sqrt{8 .5 \times 10^{- 45}} = \sqrt{85 \times 10^{- 46}} = \sqrt{85} \times 10^{- 23} = 9 .21 \times 10^{- 23} For {Ag}_{2} {CO}_{3}, solubility can be calculated as {Ag}_{2} {CO}_{3} ⇌ {\underset{2 S}{2 Ag}}^{+} + \underset{S}{{CO}_{3}^{2 -}} K_{sp} = (2 S)^{2} \cdot S = 4 S^{3} \Rightarrow S = \sqrt[3]{\frac{K_{sp}}{4}} S = \sqrt[3]{\frac{6 .2 \times 10^{- 12}}{4}} = \sqrt[3]{1 .55 \times 10^{- 12}} = 1 .15 \times 10^{- 4}$
For existence of Pb₉IO₃, there must be ${IO}_{3}^{-} and {Pb}_{9}^{+}$ present as constituent ions but in real ${Pb}_{9}^{4} -$ exist as ion. When we match the K_sp value of iodate of lead, we found that compound must be $Pb {({IO}_{3})}_{2} .$
Now, we can calculate solubiliry of compound as
$Pb {({IO}_{3})}_{2} ⇌ {Pb}_{S}^{+} + 2 {IO}_{3}^{-} K_{sp} = S \cdot (2 S)^{2} = 4 S^{3}$
$S = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{2 .6}{4} \times 10^{- 13}} = \sqrt[3]{\frac{260}{4} \times 10^{- 15}} = \sqrt[3]{65 \times 10^{- 5}} = 4 .02 \times 10^{- 5}$
Hence, maximum solubility has been observed for (c), i.e.
${Ag}_{2} {CO}_{3}$

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