I am attempting to determine whether there is going to be the formation of a $\ce{AgCl}$ salt precipitate if the following two solutions were to be added together:

  1. $50mL$ of $\ce{NaCl}$, $0.04M$,
  2. $50mL$ of $\ce{AgNO3}$, $0.05M$.

Knowing the $K_{sp}$ of $\ce{AgCl}$ is $1.6 \times 10^{-10}$.

My thought process is the following:

  • Determine the reaction that takes place: $\ce{NaCl + AgNO3 -> AgCl + NaNO3}$ (is this an equilibrium? How could I certainly say?),
  • Determine the amount of $\ce{AgCl}$ that forms at the end of the reaction - noticing that $\ce{NaCl}$ limits the reaction and that, thus, there are going to be at most $n$ moles of $\ce{Cl-}$. I would say the amount of $\ce{AgCl}$ that forms, at most, is $50mL \cdot 0.04M = 0.002mol$,

And at this point I'm stuck. If the solubility product, $K_{sp}$ of $\ce{AgCl}$, expresses the amount of ions that can be dissolved in the solution at the equilibrium, I would be able to say that, since the amount of silver chloride that forms ($0.02M$, in $100mL$) is greater than the value of $s$:

$$s = \sqrt{K_{sp}} = 1.265 \times 10^{-5}M = \ce{[Ag+]_{eq} = [Cl-]_{eq} = [AgCl]_{eq}}$$

from $K_{sp} = s \cdot s$, since there are going to be equal amounts of $\ce{Ag+}$ and $\ce{Cl-}$ ions dissolved in the solution, then I would assume that some $\ce{AgCl}$ would precipitate out: $\ce{AgCl}_{precipitate} = 0.002mol - 1.265 \times 10^{-5}mol = 0,0019987$, or $(107.9u + 35.45u) \cdot 0,0019987 = 0.286g$.

Are my assumptions on the solubility product correct?

Also, if I wanted to calculate the molar concentration of the ionic species ($\ce{Ag+}$ and $\ce{Cl-}$) in the saturated solution... wouldn't I take the previously found value of $s$ and divide it by the total volume of the solution ($100mL$)? That would be $1.265 \times 10^{-5}M$ for both $\ce{Ag+}$ and $\ce{Cl-}$?

Thank you!

  • 2
    $\begingroup$ Instead of $50mL$ ( $50mL$ ), write $\pu{ 50 mL}$ ( $\pu{ 50 mL}$ }. As typographically, there is space between the value and unit, with the latter never in italic. Similarly for M and mol. $\endgroup$
    – Poutnik
    Sep 13, 2022 at 22:29

2 Answers 2


Here is a summary of the steps I would use:

  1. Determine the concentration of silver and chloride ions in the mixture, assuming that no silver chloride precipitates.
  2. Compare the solubility product to the product of these concentrations.
  3. If point 2. suggests precipitation, determine the equilibrium concentration of ions if requested.

The concentrations of silver and chloride ions are $\pu{0.025 mol L-1}$ and $\pu{0.02 mol L-1}$, respectively. Dropping the unit $\pu{mol L-1}$ and multiplying the values gives a product larger than the solubility product, so the solution is supersaturated and silver chloride will precipitate as the system attains equilibrium. At equilibrium, the following holds:

$$(0.025 - x)(0.02 - x) = \pu{1.6E−10} = K_{sp},$$

where $(0.025 - x) \pu{mol L-1}$ is the remaining concentration of silver ions, and $(0.020 - x) \pu{mol L-1}$ the remaining concentration of chloride ions. The silver ions are in excess by $\pu{0.005 mol L-1}$; because they form a 1:1 salt, this excess will remain as silver chloride precipitates.

If you try to solve this quadratic equation the usual way, you might run into numerical problems. Instead, we can substitute $$ y = 0.02 - x$$ and solve the equation below instead:

$$(0.005 + y)(y) = \pu{1.6E−10} = K_{sp},$$

where $(0.005 + y) \pu{mol L-1}$ is the remaining concentration of silver ions, and $y\ \pu{mol L-1}$ the remaining concentration of chloride ions. To avoid the quadratic equation, estimating $(0.005 + y)$ as 0.005 gives a solution of:

$$y \approx \frac{\pu{1.6E−10}}{0.005} = \pu{3.2E-8},$$

very close to the physically sensible solution of the quadratic equation.

So the remaining concentration of chloride ions is about $\pu{3.2E-8 mol L-1}$ and the remaining concentration of silver ions is about $\pu{0.005 mol L-1}$.

This calculation assumed that the given values are exact, and no corrections for non-ideality (i.e. activity coefficients different from 1, volumes of solutions not additive, etc.) were attempted.


First let's calculate the initial amounts of both reactants, $\ce{Ag+}$ and $\ce{Cl^-}$. They are :

$\ce{n(Cl^-) = c·V = 0.05 L·0.04 M = 2·10^{-3} mol}$

$\ce{n(Ag+) = c·V = 0.05 L · 0.05 M = 2.5·10^{-3} mol}$

As a consequence, the ions $\ce{Ag^+}$ are in excess (and the excess is n($\ce{Ag+) = 0.5·10^{-3}}$ mol). So the amount of $\ce{AgCl}$ is probably equal to the initial amount of $\ce{Cl-}$, and it is $2· 10^{-3}$ mol. But it is necessary to know the level of the error on n$\ce{(AgCl})$ before going further away, and taking the solubility product into account.

Of course, n($\ce{AgCl}$) has a yet unknown uncertainty. But let's start from the beginning and assume that the minimum uncertainty on the initial concentration of $\ce{NaCl}$ is $0.04$ M ± $0.001$ M. This makes sense as no figures are given after the figure $4$ included in the value $0.04$ M. This absolute error (or uncertainty) corresponds to a relative uncertainty on the initial n$\ce{(NaCl})$ and on the final n($\ce{AgCl})$ which is equal or bigger than $\pu{\frac{0.001}{0.04} = 2.5}$%. As a consequence, the absolute uncertainty on the initial amount of $\ce{NaCl}$ (and on the final amount of $\ce{AgCl}$) is equal or greater than $\ce{0.025·2·10^{-3} mol = 5·10^{-5}}$ mol. So the final amount of $\ce{AgCl}$ is $\ce{n(AgCl) = (2 ± 0.05)·10^{-3}}$ mol.

Please note that this value is provisory, because it has been obtained without taking the $\ce{AgCl}$ solubility product into account. Let's do this calculation now.

As the solubility product of $\ce{AgCl}$ is $1.6· 10^{-10}$ $\ce{M^2}$, the remaining concentrations in the final solution are :

$\ce{[Ag^+] = n(excess)/V = \frac{0.5·10^{-3} mol}{0.1 L}= 5 ·10^{-3}}$ M

$\ce{[Cl^-] = \frac{K_s}{[Ag^+]} = \frac{1.6·10^{-10}}{5·10^{-3}} = 3.2·10^{-8} }$ M

The corresponding amounts of $\ce{Ag+}$ and $\ce{Cl-}$ remaining in the final solution are then :

n$(\ce{Ag^+) = [Ag^+]·V = 5·10^{-3} M·0.1 L = 5·10^{-4}}$mol

n($\ce{Cl^-) = [Cl-]·V = 3.2 10^{-8} M·0.1 L = 3.2·10^{-9}}$ mol

As this amount of $\ce{Cl^-}$ is remaining in solution, it means that $3.2·10^{-9}$ mole of $\ce{AgCl}$ should be removed from the amount of $\ce{AgCl}$ supposed to be precipitated previously : $(2 ± 0.05)·10^{-3}$ mol. So the exact final amount of $\ce{AgCl}$ is :

$\ce{n(AgCl) = (2 ± 0.05)·10^{-3} \mathrm{mol} - 3.2·10^{-9}}$ mol = $\ce{(2.00 ± 0.05)·10^{-3}}$ mol.

This long demonstration shows that the effect of the solubility product is smaller than the final errors. So it is negligible.

  • $\begingroup$ @Karsten. Thank you for the message. I have edited the correction. $\endgroup$
    – Maurice
    Sep 16, 2022 at 8:30

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