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Ion Electron Method

What mass of ...

Question

What mass of AgI will dissolve in $1.0 L$ of $1.0 M N H_{3}$ ? Neglect change in conc. of $N H_{3}$

[Given : $K_{s p} (A g I) = 1.5 \times 10^{- 16}$ ; $K_{f} [A g (N H_{3})_{2}^{+}] = 1.6 \times 10^{7}]$ ; (At. wt. of Ag = 108 ; I =127)

4.9 \times 10^{- 5} g

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0.0056 g

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0.035 g

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0.011 g

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Solution

The correct option is D $0.011 g$
Given:-
1.L of $1 M {N H}_{3}$

$k s p (A g I) = 1.5 \times 10^{- 16}; k f [A g {({N H}_{3})}_{3}^{+}] = 1.6 \times 10^{7}$

$M_{A g} = 108 g; M_{I} = 127 g$
Now, $A_{g} I_{(s)} = A g^{+} (a q) + I^{-} (a q) ($ when dissolved $)$

and $A g^{+} (a q) + 2 N H_{3} (a q) = A g {(N H_{3})}_{3}^{+} (a q)! (K_{t})$

Net: $¯ ¯¯¯¯¯¯¯¯ ¯ A g I + 2 N H_{3} = A g {(N H_{3})}_{2} + + I^{-} (k_{q})$

$∴ k e q = K_{S} P \times k_{f} = 1.5 \times 10^{- 16} \times 1.6 \times 10^{7} = 2.4 \times 10^{- 9}$

also, ${A g I (s) + 2 N H_{3} (a q) = A g (N H})}_{2} + (a a) + I^{-} (a q)$
equilibenium $1 - x x$

$\begin{matrix} ∴ \frac{x^{2}}{1 - x} = 2.4 \times 10^{- 9} x^{2} + 2 \cdot 4 \times 10^{- 2} x - 2 \cdot 4 \times 10^{- 9} = 0 so, x = 4.9 \times 10^{- 5} mol / L \end{matrix}$

So, moles of $A g {({N H}_{3})}_{2} + = 4.9 \times 10^{- 5} \times 1 = 4.9 \times 10^{- 5} m o l e s$
mars of AgIrequired= $108 \times 4.9 \times 10^{- 5} + 127 \times 4.9 \times 10^{- 5}$

$\begin{matrix} = 1151.5 \times 10^{- 5} g = 0.011515 g \approx 0.01 l g \end{matrix}$

So, D) $0.011 g$ is covrect

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