For the reaction [ Ag ( CN )2]-( aq ) leftharpoons Ag +( aq )+2 CN - (aq), the equilibrium constant at 25° C is 4.0 × 10-19. Calculate the silver ion concentration in a solution which was originally 0.10 M in KCN and 0.03 M in AgNO 3.

Question

For the reaction [ Ag ( CN )2]-( aq ) leftharpoons Ag +( aq )+2 CN - (aq), the equilibrium constant at 25° C is 4.0 × 10-19. Calculate the silver ion concentration in a solution which was originally 0.10 M in KCN and 0.03 M in AgNO 3. (A) 2.5 × 10-18 M (B) 1.5 × 10-18 M (C) 5.5 × 10-18 M (D) 7.5 × 10-18 M

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/6b9682d323260e9641efe1702756d108-.png" /> K =([ Ag +][ CN -]2/[ Ag ( CN )2]-) =4 × 10-19 =( a (0.04+2 a )2/[ Ag ( CN )2]-) On solving 4 × 10-19 ≈ ( a (0.04)2/(0.03- a )) i.e., a =7.5 × 10-18 M

Q. For the reaction $[A g (CN)_{2}]^{-} (a q) ⇌ A g^{+} (a q) + 2 C N^{-} (a q)$ , the equilibrium constant at $2 5^{\circ} C$ is $4.0 \times 1 0^{- 19}$ . Calculate the silver ion concentration in a solution which was originally $0.10 M$ in $K CN$ and $0.03 M$ in $A g N O_{3}$ .

$2.5 \times 1 0^{- 18} M$

$1.5 \times 1 0^{- 18} M$

$5.5 \times 1 0^{- 18} M$

$7.5 \times 1 0^{- 18} M$

$K = \frac{[ A g ^{+} ] [ C N ^{-} ] ^{2}}{[ A g ( CN ) ^{2} ] ^{-}}$
$= 4 \times 1 0^{- 19}$
$= \frac{a ( 0.04 + 2 a ) ^{2}}{[ A g ( CN ) _{2} ] ^{-}}$
On solving $4 \times 1 0^{- 19} \approx \frac{a ( 0.04 ) ^{2}}{( 0.03 - a )}$
i.e., $a = 7.5 \times 1 0^{- 18} M$

Q. For the reaction [Ag(CN)2​]−(aq)⇌Ag+(aq)+2CN−(aq), the equilibrium constant at 25∘C is 4.0×10−19. Calculate the silver ion concentration in a solution which was originally 0.10M in KCN and 0.03M in AgNO3​.

2.5×10−18M

1.5×10−18M

5.5×10−18M

7.5×10−18M