How many moles NH 3 must be added to 2.0 litre of 0.80 M AgNO 3 in order to reduce the Ag + concentration to 5 × 1 0 − 8 M . ( K f of [ Ag ( NH 3 ) 2 + ] = 1 0 8 )

Question

How many moles NH 3 ​ must be added to 2.0 litre of 0.80 M AgNO 3 ​ in order to reduce the Ag + concentration to 5 × 1 0 − 8 M . ( K f ​ of [ Ag ( NH 3 ​ ) 2 ​ + ] = 1 0 8 )

TestHub · Accepted Answer

Here's a concise solution:

The reaction is: $A g^{+} + 2 N H_{3} ⇌ [A g (N H_{3})_{2}]^{+}$ with $K_{f} = 1 0^{8}$ .

Let $x$ be the moles of $N H_{3}$ added. The concentration of $A g^{+}$ is $5 \times 1 0^{- 8} M$ . Initial moles of $A g N O_{3}$ is $2.0 L \times 0.80 M = 1.6$ moles.

$K_{f} = \frac{[[ A g ( N H _{3} ) _{2} ] ^{+} ]}{[ A g ^{+} ] [ N H _{3} ] ^{2}} = 1 0^{8}$

$1 0^{8} = \frac{1.6/2}{( 5 \times 1 0 ^{- 8} ) ( x /2 ) ^{2}}$

$1 0^{8} = \frac{0.8}{( 5 \times 1 0 ^{- 8} ) ( x ^{2} /4 )}$

$x^{2} = \frac{0.8 \times 4}{5 \times 1 0 ^{- 8} \times 1 0 ^{8}} = \frac{3.2}{5} = 0.64$

$x = 0.64 = 0.8$

Moles of $N H_{3}$ = $x$ = $0.8 M \times 2 L$ = 4 moles.

So, the answer is 4.

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