SATHEE: Equilibrium Question 913

Equilibrium Question 913

Question: The number of moles of $N H_{3}$ that must be added to 1 L of 0.1 M $A g N O_{3}$ to reduce $A g^{+}$ concentration to $2 \times 10^{- 7} M$ are Given, $K_{d i s} {[A g {(N H_{3})}_{2}]}^{+} = 6.8 \times 10^{- 8}$

Options:

A) 0.184 M

B) 0.384 M

C) 0.293 M

D) 0.0539 M

Show Answer

Answer:

Correct Answer: B

Solution:

${[A g, {(N H_{3})}_{2}]}^{+} = 0.1, M$

$[A g^{+}] = 2 \times 10^{- 7}$

${[A g {(N H_{3})}_{2}]}^{+} A g^{+} + 2 N H_{3}$

$K_{d i s} = \frac{[A g^{+}] {[N H_{3}]}^{2}}{{[A g {(N H_{3})}_{2}]}^{+}}$

$6.8 \times 10^{- 8} = \frac{2 \times 10^{- 7} \times {[N H_{3}]}^{2}}{0.1}$

$[N H_{3}] = 0.184, M$

${[N H_{3}]}_{t o t a l} = {[N H_{3}]}_{f r e e} + {[N H_{3}]}_{c o m p l e x e d}$

$= 0.184 + 2 \times 0.1 = 0.384, M$

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Equilibrium Question 913

Question: The number of moles of NH3 that must be added to 1 L of 0.1 M AgNO3 to reduce Ag+ concentration to 2×10−7M are Given, Kdis[Ag(NH3)2]+=6.8×10−8

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Question: The number of moles of $N H_{3}$ that must be added to 1 L of 0.1 M $A g N O_{3}$ to reduce $A g^{+}$ concentration to $2 \times 10^{- 7} M$ are Given, $K_{d i s} {[A g {(N H_{3})}_{2}]}^{+} = 6.8 \times 10^{- 8}$