# An AgNO_3 solution (44 ml/0.44 M) has been mixed with an NaCl solution (83 ml/1.35 times 10^{-2}...

## Question:

An {eq}\rm AgNO_3 {/eq} solution {eq}\rm (44\ ml/0.44\ M) {/eq} has been mixed with an {eq}\rm NaCl {/eq} solution {eq}\rm (83\ ml/1.35 \times 10^{-2}\ M) {/eq}.

a) Determine the ion product of the precipitate(potential precipitate).

b) Would a precipitate be created? The {eq}\rm K_{SP} {/eq} of {eq}\rm AgCl(s) {/eq} proves to be {eq}\rm 1.8 \times 10^{-10} {/eq}.

## Solubility Product Constant

The solubility product constant {eq}K_{sp} {/eq} is an equilibrium constant describing the dissociation of an insoluble salt.

Consider, for example, the dissociation of lead chloride {eq}\text{PbCl}_2 {/eq}:

{eq}\begin{align*} \text{PbCl}_2\text{(s)} \rightleftharpoons \text{Pb}^{2+}\text{(aq)} + \text{Cl}^-\text{(aq)} \end{align*} {/eq}

The {eq}K_{sp} {/eq} expression, which is just the equilibrium constant of the dissociation reaction, is given by:

{eq}\begin{align*} K_{sp} &= [\text{Pb}^{2+} ][\text{Cl}^- ]^2 \end{align*} {/eq}

Note that solids are excluded from the equilibrium expression.