Solubility Product

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Homework Statement



What is the minimum quantity of ammonia which must be added to 1L of a solution in order to dissolve 0.1 mole silver chloride by forming [Ag(NH3)2]+.
Given Ksp- 10^-10 and Kf- 10^8


Homework Equations





The Attempt at a Solution



I used the equation (which I derived) s=c^2/(Ksp)(Kf) (assuming almost all amount of AgCl converts to complex and some other approximations)
where s is the solubility of AgCl and c is the concentration of NH3 (both in mol/litre)

I solved and got c as 1 mole.
But answer given is 1.2 mole.

Are my assumptions wrong, or there is some other thing?
 

Answers and Replies

  • #2
Borek
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Show your derivation.
 
  • #3
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Let s be the solubility of AgCl in cM NH3. Let x mole/L be the amount of salt forming complex

At equilibrium.
Ag+ = s-x
ksp=(s-x)(s)
kf= x/(s-x)(c-2x)^2
Since kf is very high, x approaches s. Thus s-x would be very small. Let this value be y. Since x is small, 2x can be ignored w.r.t c.

i.e. c-2x = c
ksp=ys
kf=s^2/ksp(c)^2
i.e.
s=c x (ksp x kf)^1/2

Sorry, I wrote the earlier expression wrong.
 
  • #4
Borek
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You want x to be around 0.1M. It is not small.
 
  • #5
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Oops. Then that will make the calculations very lengthy.
Is there any method by which we solve this way and then add 0.2 mole by giving some reason?
 
  • #6
Borek
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What is [Cl-]?

What is maximum possible [Ag+]?

What is complex concentration?

Calculate ammonia concentration that fits.

--
 
  • #7
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max [CL-] is 0.1
max [Ag+] is 10^-9

How to calculate ammonia concentratrion that fits?
 
  • #8
Borek
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What is complex concentration?

Look at Kf.
 
  • #9
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Complex concentration will be practically 0.1.
From this
[NH3]^2 = 0.1/(10^-9)(10^8)
I got [NH3] as 1M
not 1.2.
 

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