A silver wire dipped in 0-1M HCI solution saturated with AgCI develops a potential of -0-25V- If E0Ag-Ag-0-799 V- the solubility of AgCl in pure water will be -

The correct option is A 7.16 × 10^ 6 mole/litre E_Ag|AgCl|Cl^ =E_Ag|Ag^+^0+0.0591/1log[Cl^ ]/[K_sp] 0.25 = 0.799 + 0.0591 log0.1/K_sp K ...

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Byju's Answer

Standard XII

Chemistry

Equilibrium Constant from Nernst Equation

A silver wire...

Question

A silver wire dipped in 0.1M HCI solution saturated with AgCI develops a potential of -0.25V. If $E_{A g / A g +}^{0} = - 0.799 V$ , the solubility of AgCl in pure water will be :

7.16 \times 10^{- 6} m o l e / l i t r e

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5.5 \times 10^{- 6} m o l e / l i t r e

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6.28 \times 10^{- 6} m o l e / l i t r e

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None of the above

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Solution

The correct option is A $7.16 \times 10^{- 6} m o l e / l i t r e$
$E_{A g | A g C l | C l^{-}} = E_{A g | A g^{+}}^{0} + \frac{0.0591}{1} l o g \frac{[C l^{-}]}{[K_{s p}]}$
$⟹ - 0.25 = - 0.799 + 0.0591 l o g \frac{0.1}{K_{s p}}$

$⟹ K_{s p} = 5.136 \times 10^{- 11}$
$∴$ Solublity $= \sqrt{K_{s p}} = 7.16 \times 10^{- 6}$ mol/litre.

Option A is correct.

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A silver wire dipped in 0.1M HCI solution saturated with AgCI develops a potential of -0.25V. If E0Ag/Ag+=−0.799V, the solubility of AgCl in pure water will be :

The correct option is A 7.16×10−6mole/litreEAg|AgCl|Cl−=E0Ag|Ag++0.05911log[Cl−][Ksp]⟹−0.25=−0.799+0.0591log0.1Ksp⟹Ksp=5.136×10−11∴ Solublity =√Ksp=7.16×10−6mol/litre.Option A is correct.

A silver wire dipped in 0.1M HCI solution saturated with AgCI develops a potential of -0.25V. If $E_{A g / A g +}^{0} = - 0.799 V$ , the solubility of AgCl in pure water will be :