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Calculate the solubility of silver chloride. a) in pure water b) Calculate the minimum concentration of ammonia that prevents precipitation in a solution that contains $0.1$ mol of AgNO3 and $0.01$ mol of NaCl per litre. Data: Kf [Ag(NH3)2+] = $1.6×10^7$ Ksp(AgCl)= $1.8×10^{−10}$

For part a) I'm totally sure on how do it: $$AgCl\rightarrow Ag^+ +Cl^-$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ s \ \ \ \ \ \ s$$ $K_s=s^2\implies s=\sqrt{K_s}$

But for part b) I'm really stuck, I've tried $$\ce{Ag^+ +2NH3 \rightarrow Ag(NH3)2}$$ and $$\ce{AgCl\rightarrow Ag+ +Cl-}$$ but not sure where to go from here.

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    $\begingroup$ Hint: $\ce{[Cl-]} = \pu{0.01 mol}$, when you only consider one liter of the solution $\endgroup$ Oct 25, 2022 at 12:24
  • $\begingroup$ yes so then we can find the $[Ag^+]$ with the constant of solubility which gives $1.8*10^{-8}$ but then I don't know what to do $\endgroup$
    – Aley20
    Oct 25, 2022 at 12:27
  • $\begingroup$ You also know Kf, can you do something with that? $\endgroup$ Oct 25, 2022 at 13:34
  • $\begingroup$ @SafdarFaisal something like this maybe? $K_f=\frac{1.8*10^{-8}x}{x}$ but I'm unsure since we didn't know the moles of $\ce{Ag(NH3)2}$ either $\endgroup$
    – Aley20
    Oct 25, 2022 at 15:36

1 Answer 1

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In pure water, you have probably seen that $s \ce{= 1.34·10^{-5} M}$. In a solution containing $\ce{Cl-}$ ions in concentration $0.01$ M, the concentration in $\ce{Ag+}$ is then :

$$[\ce{Ag+}] = \frac{K_\mathrm{s}}{[\ce{Cl^-}]} = \frac{1.8 \times 10^{-10}}{0.01} = 1.8 \times 10^{-8}$$

The definition of $K_\mathrm{f}\ce{(Ag(NH3)2^+)}$ is

$$K_f = \frac{[\ce{Ag(NH3)2^+}]}{[\ce{Ag^+}][\ce{NH3}]^2} = 1.6\times 10^7$$

so that the critical concentration of $\ce{NH3}$ is given by:

$$[\ce{NH3}]^2 = \frac{[\ce{Ag(NH3)2^+}]}{[\ce{Ag+}]\cdot K_\mathrm{f}} = \frac{0.1}{1.8\times 10^{-8}\times 1.6\times10^7} = 0.347$$

The final $\ce{NH3}$ concentration is $\sqrt{0.347} = \pu{0.59 M}$

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  • $\begingroup$ The value of $K_\mathrm{s}$ given was $1.8 \times 10^{-10}$, I took the liberty to change that $\endgroup$ Oct 25, 2022 at 15:49
  • $\begingroup$ @Maurice amazing! But why is $\ce{[Ag(NH3)2+]}=0.1$? $\endgroup$
    – Aley20
    Oct 25, 2022 at 17:26
  • $\begingroup$ @ Aley20. Silver ions are present under two forms : $\ce{Ag^+}$ and $\ce{Ag(NH3)2^+}$. The total of these two ions is $0.1$ . But, as $\ce{[Ag^+] = 1.8 · 10^{-8}}$, this value is negligible with respect to $0.1$, so that $\ce{[Ag(NH3)2^+]}$ is practically equal to $0.1$ $\endgroup$
    – Maurice
    Oct 25, 2022 at 21:45
  • $\begingroup$ @Maurice oh okay! So does it always add up to 0.1 or is it only because we had 0.1M of $\ce{AgNO3}$? I'm guessing the latter but I want to be sure. $\endgroup$
    – Aley20
    Oct 26, 2022 at 7:13
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    $\begingroup$ @ Aley20. You are right. The given value ($0.1 M$), is the concentration in $\ce{AgNO3}$. $\endgroup$
    – Maurice
    Oct 26, 2022 at 15:43

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