# Using solubility product to determine the mass of the precipitate of a reaction

I mix together $$\pu{100 mL}$$ of an aqueous $$\ce{NaCl}$$ solution at $$\pu{0.5 M}$$ and $$\pu{100 mL}$$ of an aqueous $$\ce{AgNO3}$$ solution at $$\pu{0.3 M}$$. Assuming that the solubility of $$\ce{NaCl}$$ in water is $$\pu{360 g/L}$$, the solubility of $$\ce{AgNO3}$$ in water is $$\pu{2340 g/L}$$, the solubility of $$\ce{NaNO3}$$ is $$\pu{921 g/L}$$ and the $$K_\mathrm{sp}$$ of $$\ce{AgCl}$$ is $$1.77 \times 10^{-10}$$, what is the mass of the precipitate ($$\ce{AgCl}$$) formed?

I know that the answer to this problem is $$\pu{42.1 g}$$, I solved it using stoichiometry. However, I want to know how to solve it using the solubility product and the other solubility figures given.

• @EdV Apparently there is a way to use the solubility figures given to get to the same answer and I would like to know how. Jun 4 '20 at 0:38
• @EdV well, I was told to use the notions of equilibrium solubility, which is not what I used to find 42.1 g Jun 4 '20 at 0:42
• What is the shaggy dog ? Jun 4 '20 at 10:17
• @EdV: With all the respect, the answer is $\pu{4.3 g}$. I think, you forgot the initial volume of $\ce{AgNO3}$. :-) Jun 4 '20 at 14:30
• @MathewMahindaratne You are correct: I spaced a factor of 10 due to the volume of silver nitrate solution. I will delete some comments as penance and thanks for correcting me! ;-)
– Ed V
Jun 4 '20 at 15:02

This is awfully closed to a homework question. Since you already have an answer (real answer is actually $$\pu{4.3 g}$$, see following calculations) and still insisting on how to use the solubility product information, I'd like to give some insight. Since all components are in solutions, solubilities of $$\ce{AgNO3}$$, $$\ce{NaCl}$$, and $$\ce{NaNO3}$$ should not be in concern. Solubility of $$\ce{AgCl}$$ is the only one should be concern:

$$\begin{array}{lccc} \ce{&2 AgNO3(aq) & + NaCl(aq) &<=> &AgCl(s) &+ & NaNO3(aq)} \\ \text{Initial}, \pu{mol} & 0.3 \times 0.10 = 0.03 & 0.5 \times 0.10 = 0.05 && 0 && 0 \\ \text{Final}, \pu{mol} & 0 & 0.05 - 0.03 = 0.02 && 0.03 && 0.03 \\ \text{Final}, \pu{mol/L}& 0 & \frac{0.02}{0.2}= 0.1 && \pu{0.03 mol} && \frac{0.03}{0.2}= 0.15 \\ \end{array}$$

Thus, amount of $$\ce{AgCl}$$ precipitated $$= \pu{0.03 mol} \times \pu{143.3 g/mol} = \pu{4.3 g}$$

Since $$K_\mathrm{sp}^\ce{AgCl} = \pu{1.77 \times 10^{-10} mol^2L^-2}$$ is such a small number, it is safe to say almost all of $$\ce{AgCl}$$ is in solid form. However, the solubility product come to play a role if there is a question about what is the final $$\ce{[Ag+]}$$ in solution. Now you have to consider following equilibrium:

$$\ce{AgCl(s) + H2O <=> Ag+ + Cl-}$$

Hence, $$K_\mathrm{sp}^\ce{AgCl} = [\ce{Ag+}][\ce{Cl}] = \pu{1.77 \times 10^{-10} mol^2L^-2} \tag1$$

Suppose the solubility of $$[\ce{Ag+}]$$ is $$\alpha \ \pu{ mol\:L-1}$$. Then, $$[\ce{Cl-}]$$ is $$\alpha \ \pu{+ 0.1 mol\:L-1}$$ (common ion effect). Yet, $$K_\mathrm{sp}^\ce{AgCl} = \pu{1.77 \times 10^{-10} mol^2L^-2}$$ is such a small number, the $$\alpha$$ amount from the dissolution of $$\ce{AgCl}$$ can be considered negligible.

Thus, from equation $$(1)$$:

$$K_\mathrm{sp}^\ce{AgCl} = [\ce{Ag+}][\pu{0.1 mol L-1}] = \pu{1.77 \times 10^{-10} mol^2L^-2} \\ \Rightarrow [\ce{Ag+}] = \frac{\pu{1.77 \times 10^{-10} mol^2L^-2}}{\pu{0.1 mol L-1}} = \pu{1.77 \times 10^{-9} molL-1}$$