# Question be392

Jun 9, 2015

(a) The sample contained 0.002306 moles of chloride ions;
(b) The sample contained 0.0817 g of chloride ions.

#### Explanation:

The idea behind this reaction is that the silver nitrate, $A g N {O}_{3}$, will react with the chloride ions, $C {l}^{-}$, to form silver chloride, $A g C l$, a white precipitate.

In order to determine whether or not all the chloride ions have been consumed by the reaction, you add chromate ions, $C r {O}_{4}^{2 -}$, usually in the form of potassium chromate,as an indicator. The idea is that the chromate ions will react with the silver cations and form a red precipitate only if no more chloride can be found in the solution.

So, the net ionic equation for the reaction between the chloride ions and silver nitrate looks like this

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

Notice that you have a $1 : 1$ mole ratio between the silver cations and the chloride anions. This means that, in order for a reaction to take place, you need equal numbers of moles of each ion present in solution.

In your case, the end point of the reaction required 43.75 mL of a 0.05270-M silver nitrate solution. Use this information to determine how many moles of $A {g}^{+}$ were needed to precipitate all the chloride ions.

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{A {g}^{+}} = \text{0.05270 M" * 43.75* 10^(-3)"L" = "0.002306 moles}$ $A {g}^{+}$

According to the aforementioned mole ratio, you have

$0.002306 \cancel{\text{moles"Ag^(+)) * "1 mole Cl"^(-)/(1cancel("mole"Ag^(+))) = color(green)("0.002306 moles } C {l}^{-}}$

To determine how many grams of chloride ions you had present in the sample, use chloride's molar mass

0.002306cancel("moles") * "35.45 g"/(1cancel("mole")) = color(green)("0.08175 g "Cl^(-)#

SIDE NOTE Looking at the mass of the sample you started with, I assume that you'll have to calculate the mass percentage of chloride ions in the original sample.

To do that, simply divide the mass of the chloride ions by the total mass of the sample and multiply by 100.