If silver nitrate is added to an aqueous solution (such as tap water) that contains both chloride ions and phosphate ions, which precipitate is more likely to form: silver chloride, silver phosphate or both should form? And how do you know?

This is NOT a homework question. I'm 32 and learning chemistry on my own as a hobby and doing experiments using a Chemistry Kit for kids. The reason I'm asking the question is because I did the experiment on the tap water at home and I didn't get any silver phosphate precipitate, but I did get a milky white appearance which I'm assuming is silver chloride (or another halide). I'm wondering if I got no silver phosphate precipitates because all the silver ions were "used up" by the chloride ions or simply because the concentration of phosphate ions is either zero or too small.

I tried to figure out if silver chloride formation takes precedence (so to speak) over forming silver phosphates using enthalpies of formation, but I got lost with that a bit.

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    $\begingroup$ Depends on the concentrations of the ions and the solubility products of the precipitates. $\endgroup$ Sep 2, 2017 at 20:25

1 Answer 1


The key concept here is solubility products which are a form of chemical equilibria. Once you have understood that, you will realise that it depends on the ions’ concentrations and relative solubilities which one will precipitate first.

In general, you have to consider the following principal equilibria:

$$\ce{Ag3PO4 v + 3 Cl- (aq) <=>[$k_1$] 3 Ag+ (aq) + 3 Cl- (aq) + PO4^3- <=>[$k_2$] 3 AgCl v + PO4^3-}\tag{1}$$

and futher ones that derive from the acidity equilibrium of the phosphate anions as well as the formation of silver hydroxide at higher $\mathrm{pH}$ values. I will choose to ignore these.

Equilibrium $k_1$ is actually only the precipitation of silver phosphate; we might instead write it as:

$$\ce{3 Ag+ (aq) + PO4^3- (aq) <=>[$k_1$] Ag3PO4 v}\tag{2}$$

The equilibrium constant is thus:

$$K_1 = \frac{a(\ce{Ag3PO4})}{a(\ce{Ag+})^3 \times a(\ce{PO4^3-})}\tag{3}$$

We are dealing with activities here; the activity of a solid is, by definition, always $1$. This equilibrium is usually given in the inverse form in which the dissolved ions are the product and the solid salt is the reactant; in that case $K_1{}^{-1}$ is what you might know as the solubility product.

The same discussion can be made for the other equilibrium:

$$\begin{gather}\ce{Ag+ (aq) + Cl- (aq) <=> AgCl v}\tag{4}\\[1em] K_2 = \frac{a(\ce{AgCl})}{a(\ce{Ag+}) \times a(\ce{Cl-})}\tag{5}\end{gather}$$

Again, the more common form is $K_2{}^{-1}$ which is also known as the solubility constant.

Knowing the values of both $K$ as well as the initial concentrations of all ions will allow you to calculate in which direction the reaction goes although the calculation may be tedious.


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