A saturated solution of Agcl is treated with solid Nacl until the [Cl-] is 0.27M.


Screenshot 2024 01 25 225104

Screenshot 2024 01 25 225104

Step 1 of 2

Given data: A saturated solution of AgCl is given, [Cl−]= 0.27M.

Find: find Ag+ in solution and its percentage.

Explanation:

To determine the resulting Ag+ concentration and the percentage of Ag+ remaining in solution, we need to consider the solubility equilibrium of AgCl.

Step 2 of 2

PART A ) AgCl is a sparingly soluble salt.

Its solubility product Ksp = 1.8×10−10

AgCl(s)=Ag+(aq)+Cl−(aq)

Ksp = S x S

Ksp = S2

1.8 x 10−10=S2

S=1.8×10−10

S=[Ag+]=1.34 x 10−5M

Thus, [Ag+]=1.34 x 10−5M in the original saturated solution.

On addition of NaCl,
[Cl−]=0.27M,

Thus, Ksp=[ Ag+] [Cl−]
1.8x 10−10=[Ag+] [ 0.27]

[Ag+]=1.8×10−100.27

[Ag+]=6.67×10−10

PART B)

%[Ag+] remains in solution=([Ag+] after addition of NaCl[Ag+] before addition of NaCl)x100%[Ag+] remains in solution=(6.67×10−101.34×10−5)x100%[Ag+] remains in solution=4.978 x 10−3%[Ag+] remains in solution=0.005%

Explanation:

In a saturated solution, the concentration of Ag+ is determined by the solubility product constant Ksp of AgCl. The Ksp expression for AgCl can be written as:

Ksp = [Ag+][Cl-]

Final solution

The resulting Ag+ in solution is = 6.67×10−10

%[Ag+] remains in solution= 0.005%

 

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