# Find the minimum concentration so that it doesn't precipitate

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.

• Hint: $\ce{[Cl-]} = \pu{0.01 mol}$, when you only consider one liter of the solution Commented Oct 25, 2022 at 12:24
• 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 Commented Oct 25, 2022 at 12:27
• You also know Kf, can you do something with that? Commented Oct 25, 2022 at 13:34
• @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 Commented Oct 25, 2022 at 15:36

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 $$\sqrt{0.347} = \pu{0.59 M}$$

• The value of $K_\mathrm{s}$ given was $1.8 \times 10^{-10}$, I took the liberty to change that Commented Oct 25, 2022 at 15:49
• @Maurice amazing! But why is $\ce{[Ag(NH3)2+]}=0.1$? Commented Oct 25, 2022 at 17:26
• @ 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 $0.1$ Commented Oct 25, 2022 at 21:45
• @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. Commented Oct 26, 2022 at 7:13
• @ Aley20. You are right. The given value ($0.1 M$), is the concentration in $\ce{AgNO3}$. Commented Oct 26, 2022 at 15:43