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Equilibrium Constant from Nernst Equation

Excess of A...

Question

Excess of $A g C l$ is added to $0.1 M$ solution of $K B r$ at $298 K$ . Calculate the equilibrium concentrations of $B r^{-}$ and $C l^{-}$ ions.
$E_{C l^{-} / A g C l / A g}^{\circ} = 0.222 v o l t E_{B r^{-} / A g B r / A g}^{\circ} = 0.095 v o l t$ .

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Solution

$[C l^{-}] = 0.09929 m o l L^{- 1}, [B r^{-}] = 0.00071 m o l L^{- 1}$
$A g C l + B r^{-} (0.1 - x) ⇌ A g B r + C l^{-} x$
Apply $log K = \frac{E_{c e l l}^{\circ}}{0.0591}$
or $log \frac{x}{(0.1 - x)} = \frac{E_{c e l l}^{\circ}}{0.0591}$ .

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Similar questions

K_{s p}

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{A g}_{3} A s O_{4}

5 \times 10^{- 13}, 1.8 \times 10^{- 10}, 8.1 \times 10^{- 12}

10^{- 22}

respectively. Silver nitrate is added to the solution each having

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concentration of

{B r}^{-}, {C l}^{-}, {C O}_{3}^{2 -}

A s O_{4}^{3 -}

. Select the correct answer.

Q. Consider the cell

A g (s) | A g B r (x) | B r^{-} (a q) | | C l^{-} (a q) | A g C l (s) | A g (s)

^{o}

C. The solubility product constants of

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Q. Consider the cell

A g | A g B r (s) B r^{-} | | A g C l (s) C l^{-} | A g

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C. The solubility product of AgCl and AgBr are

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Q. A solution is 0.1 M in each

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K B r

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A g N O_{3}

is gradually added. Assuming that there is no change in the volume of the solution given, The concentration (in M) of first ion when the second ion starts precipitating is about .

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1.7 \times 10^{- 10}

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Q. A solution of

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Given:

K_{s p} (A g C l) = 1.7 \times 10^{- 10}

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What will be the conc. of both ions when the third ion start precipitating.

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