# Solubility involving precipitation

Question
How do I found out how much $\ce{AgNO3}$ can be dissolved in 1 liter of solvent without precipitation of $\ce{AgCl}$. Solvent has $$M_{\ce{\,Cl-}} = 0.1\,{\rm M}\\M_{\ce{\,NH3}} = {\rm 1.00\,M}$$ Other given values are $$K_{\text{sp}_{\Large\ce{AgCl(s)}}} = 1.8\times 10^{-10}\\K_{\mathrm{f}_{\Large\ce{[Ag(NH3)2]+}}} = 1.6 \times 10^7$$ Answer: $4.4\rm\,g$

My effort:
Firstly I tried to find out $\ce{[Ag]}$ in saturated solution from eq.

$$K_{sp_{\Large\ce{AgCl}}} = \ce{[Ag][Cl]} 1.8 \times 10^{-10} = [\ce{Ag}] \times 0.1{\rm\,M}\times [\ce{Ag}] = 1.8 \times 10 -9$$

Then I placed that value in formation eq.

$$K_f = \frac{\ce{[Ag(NH3)2]+}}{\ce{[Ag] * [NH3]^2}}$$

$$1.6 \times 10^7 = \frac{x}{(1.8\times10^{-9} \times 1.0{\rm\,M}-2)}$$

$$x = 0.0288\,\rm M$$

and tried to find out $\ce{AgNO3}$ mass by using that by

$$m~\ce{AgNO3} = 0.0288\rm\,M \times 1.0 {\rm\,dm^3} \times 169.9 \frac{g}{mol} = 4.89\,\rm g$$

That value doesn't seem to be the answer. What did I do wrong?

• This is a homework question. We ‎have a policy which states that you should show your thoughts and/or efforts into solving the ‎problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this ‎question may get closed.‎ – M.A.R. Nov 28 '15 at 18:21
• Edited. Hope this satisfies – VerifymyD Nov 28 '15 at 18:39
• OK, I retracted my close vote. Let me edit this post for formatting, and do not edit until I have done. – M.A.R. Nov 28 '15 at 18:56
• The formatting of some of the work is still weird, but I get the same answer - 4.89 g AgNO3. – MaxW Nov 29 '15 at 2:03

# Step 1: Find equilibrium concentration for Ag+

As you correctly showed in the first step, we need to find the maximum solubility equilibrium concentration of $\ce{[Ag+]_{eq}}$ that is still allowed. Your calculation is correct, the result is $$\ce{[Ag+]_{eq}} = \frac{K_\text{sp}}{\ce{[Cl- ]}} = 1.8\cdot 10^{-9}$$

# Step 2: Find equilibrium concentration for [Ag(NH3)2]+

We now need to know how much of the $\ce{AgNO3}$ we put into the solution actually ends up in the complex $\ce{[Ag(NH3)2]+}$, which is not in danger of precipitating. The formation equilibrium is according to the reaction equation $$\ce{Ag+ + 2NH3 <=> [Ag(NH3)2]+} \,,$$ and reads $$K_\text{f} = \frac{\ce{[Ag(NH3)2+]}}{\ce{[Ag+]_{eq}}\ce{[NH3]^2}} \,.$$

Substituting $x=\ce{[Ag(NH3)2+]}$ we quickly reach $$K_\text{f} = \frac{x}{\ce{[Ag+]_{eq}}(1-2x)^2} \,.$$ This is the step where you went wrong: You also need to consider that the concentration of $\ce{NH3}$ is changed by the formation of the complex. In principle we should also have taken this into account in the section above, but there the error is many more magnitudes smaller than here.

Solving the above gives $x = \ce{[Ag(NH3)2+]_{eq}} = 0.0258942$.

# Step 3: Calculate the total amount of Ag+ in the solution

Armed with the knowledge from the two previous sections, this step is pretty trivial. As the contribution of free silver ions is many orders of magnitude smaller than the concentration of the complex, we can safely ignore it and approximate: $$\ce{[Ag+]_{tot}} = \ce{[Ag+]_{eq}} + \ce{[Ag(NH3)2+]_{eq}} \approx 0.02589$$

# Step 4: Calculate the amount of AgNO3 to add

If we assume that the activity coefficient is 1, we can directly infer the concentration $$c_{\ce{Ag^+_{tot}}} = \ce{[Ag+]_{tot}} ~ \mathrm{mol\, L^{-1}} \,.$$

From the concentration we can calculate the amount of $\ce{AgNO3}$ to add, as you did correctly, \begin{align} m_{\ce{AgNO3}} &= c_{\ce{Ag^+_{tot}}} V M_{\ce{AgNO3}} \\ &= 0.02589 ~ \mathrm{mol\, L^{-1}} \cdot 1~\mathrm{L} \cdot 169.87 ~ \mathrm{g\, mol^{-1}} \\ &= 4.398~\mathrm{g}\,, \end{align} which, after rounding to two significant digits, leads directly to the result your solution says you should have gotten:

$$m_{\ce{AgNO3}} = 4.4~\mathrm{g}$$

Tschoppi gave all the detaols, but it boils down to this: when the silver forms a comes with the ammonia, some of the ammonia is used up so that the remaining ammonia concentration is less than 1 M. The difference is enough to affect that second significant digit.