# Question 56ccc

Aug 24, 2015

For part (a) No, a precipitate will not form.

#### Explanation:

I will show you how to do point (a), and you solve point (b) as practice.

So, you're mixing two solutions that contain soluble compounds and you want to figure out if a precipitate will form.

Silver nitrate and potassium chloride will dissociate completely in aqueous solution to produce silver and potassium cations, on one hand, and nitrate and chloride anions, on the other.

If and only if the concentrations of the silver cations and chloride anions are high enough, silver chloride, an insoluble ionic compound, will precipitate out of solution.

Now, you know that silver chloride has a solubility product constant equal to 1.6 * 10^(-10. The dissociation of silver chloride in aqueous solution is represented by an equilibrium

${\text{AgCl"_text((s]) rightleftharpoons "Ag"_text((aq])^(+) + "Cl}}_{\textrm{\left(a q\right]}}^{-}$

The nitrate anions and potassium ations are spectator ions, so you can ignore them altogether.

By definition, ${K}_{s p}$ is equal to

${K}_{s p} = \left[{\text{Ag"^(+)] * ["Cl}}^{-}\right]$

This tells you that if the concentration of silver cations multiplied by the concentration of the chloride anions is at least equal to ${K}_{s p}$, a precipitate will be formed.

If these two concentrations do not satisfy this inequality

${K}_{s p} \le \left[{\text{Ag"^(+)] * ["Cl}}^{-}\right]$

then a precipitate will not be formed.

So, start by calculating the total volume of the solution

${V}_{\text{total}} = {V}_{A g N {O}_{3}} + {V}_{K C l}$

${V}_{\text{total" = "200 mL" + "900 mL" = "1100 mL}}$

Use the molarities and the initial volumes of the two solutions to determine the number of moles of each compound

$\textcolor{b l u e}{C = \frac{n}{V} \implies n = C \cdot V}$

$n = 1.0 \cdot {10}^{- 4} {\text{M" * 200 * 10^(-3)"L" = 2.0 * 10^(-5)"moles Ag}}^{+}$

and

$n = 1.0 \cdot {10}^{- 6} {\text{M" * 900 * 10^(-3)"L" = 9.0 * 10^(-7)"moles Cl}}^{-}$

The molarities of the silver cations and chloride anions in the final solution will be

["Ag"^(+)] = (2.0 * 10^(-5)"moles")/(1100 * 10^(-3)"L") = 1.82 * 10^(-5)"M"

and

["Cl"^(-)] = (9.0 * 10^(-7)"moles")/(1100 * 10^(-3)"L") = 8.18 * 10^(-7)"M"#

This means that you have

$\left[{\text{Ag"^(+)] * ["Cl}}^{-}\right] = 1.82 \cdot {10}^{- 5} \cdot 8.18 \cdot {10}^{- 7} = 1.5 \cdot {10}^{- 11}$

Since this value is smaller than ${K}_{s p} = 1.6 \cdot {10}^{- 10}$, a precipitate will not form in the final solution.