Ncert Exercises & Solutions

7.67 Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9. Determine also the molarities of individual ions.

(1) Silver chromate

${Ag}_{2} {CrO}_{4} \to 2 {Ag}^{+} + {CrO}_{4}^{2 -}$

Then,

$K_{s p} = {[A g^{+}]}^{2} [C r {O_{4}}^{2 -}]$

Let the solubility of $A g_{2} C r O_{4}$ be s.

$\Rightarrow [{Ag}^{+}] 2 s and [{CrO}_{4}^{2 -}] = s$
$Then,$
$K_{sp} = {(2 s)}^{2} . s = 4 s^{3}$
$\Rightarrow 1.1 \times 10^{- 12} = 4 s^{3}$
$. 275 \times 10^{- 12} = s^{3}$
$s = 0.65 \times 10^{- 12} = s^{3}$
$s = 0.65 \times 10^{- 4} M$

Molarity of ${Ag}^{+} = 2 s = 2 \times 0.65 \times 10^{- 4} = 1.30 \times 10^{- 4} M$

Molarity of ${CrO}_{4}^{2 -} = s = 0.65 \times 10^{- 4} M$

(2) Barium chromate:

${BaCrO}_{4} \to {Ba}^{2 +} + {CrO}_{4}^{2 -}$
$Then, K_{sp} = [{Ba}^{2 +}] [{CrO}_{4}^{2 -}]$

Let the solubility of ${BaCrO}_{4}$ . be s.

So,

$[{Ba}^{2 +}] = s and [{CrO}_{4}^{2 -}] = s \Rightarrow K_{sp} = s^{2}$
$\Rightarrow 1.2 \times 10^{- 10} = s^{2}$
$\Rightarrow s = 1.09 \times 10^{- 5} M$

Molarity of ${Ba}^{2 +}$ =Molarity of ${CrO}_{4}^{2 -} = s = 1.09 \times 10^{- 5} M$

$(3) Ferric hydroxide :$
$Fe {(O)}_{3} \to {Fe}^{2 +} + 3 {OH}^{-}$
$K_{sp} = [{Fe}^{2 +}] {[{OH}^{-}]}^{3}$
$Let s be the solubility of Fe {(OH)}_{3}$
$Thus, [{Fe}^{3 +}] = s and [{OH}^{-}] = 3 s \Rightarrow K_{sp} = s . {(3 s)}^{3} = s . 27 s^{3}$
$K_{sp} = 27 s^{4}$
$. 037 \times 10^{- 38} = s^{4}$
$. 00037 \times 10^{- 36} = s^{4} \Rightarrow 1.39 \times 10^{- 10} M = S$
$Molarity of {Fe}^{3 +} = s = 1.39 \times 10^{- 10} M$
$Molarity of {OH}^{-} = 3 s = 4.17 \times 10^{- 10} M$

$(4) Lead chloride :$
${PbCI}_{2} \to {pb}^{2 +} + 2 {CI}^{-}$
$K_{sp} = [{pb}^{2 +}] [{CI}^{-}] = 2 s$
$Thus, K_{sp} = s . {(2 s)}^{2} = 4 s^{3}$
$\Rightarrow 1.6 \times 10^{- 5} = 4 s^{3}$
$\Rightarrow 0.4 \times 10^{- 5} = s^{3}$
$4 \times 10^{- 6} = s^{3} \Rightarrow 1.58 \times 10^{- 2} M = S . I$
$Molarity of {PB}^{2 +} = s = 1.58 \times 10^{- 2} M$
$Molarity of chloride = 2 s = 3.16 \times 10^{- 2} M$
$(5) Mercurous iodide :$
${Hg}_{2} I_{2} \to {Hg}^{2 +} + 2 I^{-}$

$K_{sp} = {[{Hg}_{2}^{2 +}]}^{2} {[I^{-}]}^{2}$

Let be the solubility of ${Hg}_{2} l_{2}$ .

$\Rightarrow [{Hg}_{2}^{2 +}] = s and [l^{-}] = 2 s$
$Thus, K_{sp} = s {(2 s)}^{2} \Rightarrow K_{sp} = 4 s^{3}$
$4.5 \times 10^{- 29} = 4 s^{3}$
$1.125 \times 10^{- 29} = s^{3}$
$\Rightarrow s = 2.24 \times 10^{- 10} M$

Molarity of ${Hg}_{2}^{2 +} = s = 2.24 \times 10^{- 10} M$

Molarity of $I^{-} = 2 s = 4.48 \times 10^{- 10} M$

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