# 1.40 g of silver nitrate is dissolved in 125 mL of water. To this solution is added 5.00 mL of 1.50 M hydrochloric acid, and a precipitate forms. What is the concentration (in units of moiarity) of silver ions remaining in the solution?

Dec 10, 2015

["Ag"^(+)] = "0.00571 M"

#### Explanation:

You're dealing with a double replacement reaction in which two soluble compounds react in aqueous solution to form an insoluble solid that precipitates out of solution.

Silver nitrate, ${\text{AgNO}}_{3}$, a soluble ionic compound, will react with hydrochloric acid, $\text{HCl}$, a strong acid that dissociates completely in aqueous solution, to form the insoluble silver chloride, $\text{AgCl}$, a white precipitate.

The balanced chemical equation for this reaction looks like this

${\text{AgNO"_text(3(aq]) + "HCl"_text((aq]) -> "AgCl"_text((s]) darr + "HNO}}_{\textrm{3 \left(a q\right]}}$

The net ionic equation, which you get after removing the spectator ions, looks like this

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

Notice that the reactants take part in the reaction in a $1 : 1$ mole ratio, which means that the reaction will consume equal numbers of moles of silver nitrate and hydrochloric acid, which produce silver cations, ${\text{Ag}}^{+}$, and chloride anions, ${\text{Cl}}^{-}$, respectively.

Your goal now is to figure out how many moles of each reactant you're mixing together.

For starters, use silver nitrate's molar mass to determine how many moles you get in that $\text{1.40-g}$ sample of silver nitrate

1.40 color(red)(cancel(color(black)("g"))) * "1 mole AgNO"_3/(169.87color(red)(cancel(color(black)("g")))) = "0.008242 moles AgNO"_3

Now use the molarity and volume of the hydrochloric acid solution to determine how many moles of acid you have

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

${n}_{H C l} = \text{1.50 M" * 5.00 * 10^(-3)"L" = "0.00750 moles HCl}$

Since you have fewer moles of chloride anions, these ions will be completely consumed by the reaction. The resulting solution will contain

${n}_{A {g}^{+}} = 0.008242 - 0.00750 = {\text{0.000742 moles Ag}}^{+}$

The total volume of the solution will be

${V}_{\text{sol" = "125 mL" + "5.00 mL" = "130.0 mL}}$

The molarity of the silver cations will thus be - do not forget that the volume must be expressed in liters!

$\left[\text{Ag"^(+)] = "0.000742 moles"/(130.0 * 10^(-3)"L") = color(green)("0.00571 M}\right)$

The answer is rounded to three sig figs.