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Alexxandr [17]
3 years ago
14

since Ag and Cl are in equilibrium with AgCl, find Ksp for agcl from the concentrations. Write the expression for ksp AgCl and d

etermine its value
Chemistry
1 answer:
melamori03 [73]3 years ago
6 0

Answer:

1.69 ×10^-10

Explanation:

Given that the equation for the dissolution of AgCl in water is;

AgCl(s) ⇄Ag^+(aq) + Cl^+(aq)

Also, silver ion and chloride ion are in equilibrium with the undissociated AgCl hence we can write;

Ksp= [Ag+][Cl-]/ [AgCl]

Since the activity of the pure sold is 1, we now have;Ksp= [Ag+][Cl-]

If we know the solubility of AgCl in pure water to be 1.3 x 10^-5 M, from standard tables, and [Ag+]=[Cl-]= 1.3 x 10^-5 M = x

Then;

Ksp= x^2

Ksp= (1.3 x 10^-5)^2

Ksp= 1.69 ×10^-10

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