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Solubility Product

Find the solu...

Question

Find the solubility product of a saturated solution of $A g_{2} C r O_{4}$ in water at $298 K$ if the emf of the cell $A g | A g^{+}$ (satd. $A g_{2} C r O_{4} s o l n .) | | A g^{+} (0.1 M) | A g$ is $0.164 V$ at $298 K$ .

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Solution

$E_{c e l l} = \frac{0.059}{1} log \frac{[A g^{+}]_{R H S}}{[A g^{+}]_{L H S}}$
$0.164 = \frac{0.059}{1} log \frac{0.1}{[A g^{+}]_{L H S}}$
or $[A g^{+}]_{L H S} = 1.66 \times 10^{- 4} M$
So, $[C r O_{4}^{2 -}] = \frac{1.66 \times 10^{- 4}}{2}$
$K_{s p (A g_{2} C r O_{4})} = [A g^{+}]^{2} [C r O_{4}^{2 -}]$
$= (1.66 \times 10^{- 4})^{2} (\frac{1.66 \times 10^{- 4}}{2})$
$= 2.287 \times 10^{- 12} m o l^{3} L^{- 3}$ .

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Similar questions

The solubility product of a saturated solution of $A g_{2} C r O_{4}$ in water at 298 K will be................., if the EMF of the cell:

A g | A g^{\oplus} (s a t A g_{2} C r O_{4} s o l) | | A g (0.1 M) | A g

0.164 V

298 K

A g C l

2.6 \times 10^{- 6} S {c m}^{- 1}

298 K

λ^{\infty} ({A g}^{+}) = 63.0 S {c m}^{2} {m o l}^{- 1}

λ^{\infty} ({C l}^{-}) = 67.0 S {c m}^{2} {m o l}^{- 1}

A g (s) | A g^{+} (0.1 M) | | Z n^{2 +} (0.1 M) | Z n (s)

E_{c e l l}^{o} = - 1.562 V

K_{s p}

{A g}_{2} {C r O}_{4}

1.1 \times 10^{- 12}

298 K

{A g}_{2} {C r O}_{4}

0.1 M

A g {N O}_{3}

A g (S) A g^{+} (0.1 M) | Z n (s)

E_{c e l l}^{0} = - 1.562 V

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