A solution is a mixture of 0.05 M NaCl and 0.05 M NaI. The concentration of iodide in the solution when AgCl just starts precipitating is equal to: (Ks p AgCl =1 × 10-10 M 2 ; Ks p AgI =4 × 10-16 M 2)

Question

A solution is a mixture of 0.05 M NaCl and 0.05 M NaI. The concentration of iodide in the solution when AgCl just starts precipitating is equal to: (Ks p AgCl =1 × 10-10 M 2 ; Ks p AgI =4 × 10-16 M 2) (A) 4 × 10-6 M (B) 2 × 10-8 M (C) 2 × 10-7 M (D) 8 × 10-15 M

Tardigrade Admin · Accepted Answer

K text sp ( AgCl )>K text sp ( AgI ) Hence is AgI is precipitated before AgCl [ Ag +]to precipitated Cl -=(1 × 10-10/0.05)=2 × 10-9 M [ I -]at this stage =(K text sp ( AgI )/[ Ag +]) =(4 × 10-16/2 × 10-9)=2 × 10-7 M

Q. A solution is a mixture of $0.05 M N a Cl$ and $0.05 M N a I$ . The concentration of iodide in the solution when $A g Cl$ just starts precipitating is equal to:
$(K_{s p} A g Cl = 1 \times 1 0^{- 10} M^{2}; K_{s p} A g I = 4 \times 1 0^{- 16} M^{2})$

$4 \times 1 0^{- 6} M$

$2 \times 1 0^{- 8} M$

$2 \times 1 0^{- 7} M$

$8 \times 1 0^{- 15} M$

Q. A solution is a mixture of 0.05MNaCl and 0.05MNaI. The concentration of iodide in the solution when AgCl just starts precipitating is equal to: (Ksp​AgCl=1×10−10M2;Ksp​AgI=4×10−16M2)

4×10−6M

2×10−8M

2×10−7M

8×10−15M

Ksp ​(AgCl)>Ksp ​(AgI) Hence is AgI is precipitated before AgCl [Ag+]to precipitated Cl−=0.051×10−10​=2×10−9M [I−]at this stage =[Ag+]Ksp ​(AgI)​ =2×10−94×10−16​=2×10−7M

Q. A solution is a mixture of $0.05 M N a Cl$ and $0.05 M N a I$ . The concentration of iodide in the solution when $A g Cl$ just starts precipitating is equal to:
$(K_{s p} A g Cl = 1 \times 1 0^{- 10} M^{2}; K_{s p} A g I = 4 \times 1 0^{- 16} M^{2})$

$4 \times 1 0^{- 6} M$

$2 \times 1 0^{- 8} M$

$2 \times 1 0^{- 7} M$

$8 \times 1 0^{- 15} M$