Here it is
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No answer correct.

The soloubility product is K = [Ag+]*[Cl-]
For value Cl- = 4*10^(-5) M, the [ Ag+] = 10^(-10)/4*10^(-5) = 2,5*10(-6) M, For value Cl =  10^(-5) M, the [ Ag+] = 10^(-10)/10^(-5) = 10(-5) M.
In 0,1 l we have the 2,5* 10^(-5) mole and 10^ (-4) mol. The difference of both what is 7,5* 10^(-5) mol hast be added.

Quote from: chenbeier on January 20, 2022, 12:15:41 PM
No answer correct.

The soloubility product is K = [Ag+]*[Cl-]
For value Cl- = 4*10^(-5) M, the [ Ag+] = 10^(-10)/4*10^(-5) = 2,5*10(-6) M, For value Cl =  10^(-5) M, the [ Ag+] = 10^(-10)/10^(-5) = 10(-5) M.
In 0,1 l we have the 2,5* 10^(-5) mole and 10^ (-4) mol. The difference of both what is 7,5* 10^(-5) mol hast be added.

Your answer is wrong
Please next time check your work and spellings before you make any post on this forum.

Quote from: Pranjal Singh on January 19, 2022, 11:32:58 PM
Here it is
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In this question first we will calculate the molar solubility of AgCl solution

Concentration of Ag + and concentration of Cl- in a saturated solution can be calculated by using the equation of KSP

Ksp =[Ag+][Cl-] = 10-¹⁰
[Cl^-] = 10^-5
[Ag+] = 10^-5
In saturated solution
      Ag + + Cl- ---> AgCl
Initial 0      4*10^-5
saturated 10^-5     10^-5
solution     
It means concentration of Ag⁺ added should be =  (4*10^-5 - 1*10^-5) +1*10^-5  =4*10^-5 M

(4*10^-5 - 1*10^-5 ) this is amount needed to precipitate out extra Cl^-
1*10^-5   is the amount needed to remain in saturated solution .
However we are working in molarity and answer options are in moles
Moles = M V (L)
Work out your answer now.