# Question 1565c

Apr 22, 2015

An answer cannot be given without providing the mols of AgNO3 and the volume of the solution.

Apr 22, 2015

The molarity of the silver nitrate solution is 0.394 mM.

The balanced chemical equation for this double replacement reaction looks like this

$A g N {O}_{3 \left(a q\right)} + K C {l}_{\left(a q\right)} \to A g C {l}_{\left(s\right)} + K N {O}_{3 \left(a q\right)}$

Notice that you have a $1 : 1$ mole ratio between silver nitrate and potassium chloride, which means that you need 1 mole of silver nitrate for every 1 mole of potassium chloride in order for the reaction to take place.

The net ionic equation for this reaction is

$A {g}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-} \to A g C {l}_{\left(s\right)}$

You know the mass of the potassium chloride sample, which means that you can use the compound's molar mass to determine how many moles of the compound reacted

0.780 * 10^(-3)cancel("g") * "1 mole KCl"/(74.5513cancel("g")) = 0.0105 * 10^(-3)"moles KCl"#

This is equal to the number of moles of silver nitrate you need in order for the reaction to precipitate all the $C {l}^{-}$ ions present in solution

${n}_{A g N {O}_{3}} = {n}_{K C l} = 0.0105 \cdot {10}^{- 3} \text{moles}$

Now use the volume of the silver nitrate solution to determine its molarity

$C = \frac{n}{V} = \left(0.0105 \cdot {10}^{- 3} \text{moles")/(26.6 * 10^(-3)"L") = 3.94 * 10^(-4)"M" = color(green)("0.394 mM}\right)$