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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.

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  • $\begingroup$ @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. $\endgroup$ – agg199 Jun 4 '20 at 0:38
  • $\begingroup$ @EdV well, I was told to use the notions of equilibrium solubility, which is not what I used to find 42.1 g $\endgroup$ – agg199 Jun 4 '20 at 0:42
  • $\begingroup$ What is the shaggy dog ? $\endgroup$ – Maurice Jun 4 '20 at 10:17
  • $\begingroup$ @EdV: With all the respect, the answer is $\pu{4.3 g}$. I think, you forgot the initial volume of $\ce{AgNO3}$. :-) $\endgroup$ – Mathew Mahindaratne Jun 4 '20 at 14:30
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    $\begingroup$ @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! ;-) $\endgroup$ – Ed V Jun 4 '20 at 15:02
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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 $(1)$:

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

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