# Question 8fb51

Mar 7, 2015

ADDITIONAL INFORMATION I have a mass of 20.00g of diluted seawater (used distilled seawater for dilution) and then i titrated with $A g N {O}_{3}$ and got a titre of 23.30 mL. Now I have to find the number of moles of $A {g}^{+}$ and $C {l}^{-}$ that reacted.

So, you've performed a titration using silver nitrate, $A g N {O}_{3}$. The reaction of interest is between silver nitrate and sodium chloride, since this reaction will allow you to precipitate the chloride ions present in the seawater solution.

You'll probably have an indicator, something like potassium chromate, present as well to tell you when the silver ions have precipitated all the chloride ions.

The balanced chemical equation looks like this

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

Since you're in aqueous solution, the net ionic equation will be

"Ag_((aq))^(+) + Cl_((aq))^(-) -> AgCl_((s))#

Now, the silver ions will react with the chloride ions present in your sample in a $\text{1:1}$ mole ratio. For every 1 mole of chloride ions, you'll need 1 mole of silver ions.

Now, I assume you know the concentration of your silver nitrate solution. You will use this to determine how many moles of silver were needed to complete the reaction

$C = \frac{n}{V} \implies {n}_{A g N {O}_{3}} = {C}_{A g N {O}_{3}} \cdot {V}_{A g N {O}_{3}} = {C}_{A g N {O}_{3}} \cdot 23.30 \cdot {10}^{- 3} \text{L}$

Once you know the number moles of silver nitrate moles, you know the number of moles of silver ions, and thus the number of moles of chloride ions. Since

$A g N {O}_{3 \left(a q\right)} \to A {g}_{\left(a q\right)}^{+} + N {O}_{3 \left(a q\right)}^{-}$

the number of moles of silver ions will be equal to the number of moles of silver nitrate

${\text{? moles silver nitrate" * "1 mole silver ions"/"1 mole silver nitrate" = "? mol Ag}}^{+}$

${\text{? moles Ag"^(+) * "1 mole Cl"^(-)/"1 mole Ag"^(+) = "? moles Cl}}^{-}$

where $\text{?}$ is ${n}_{A g N {O}_{3}}$ calculated using silver nitrate's molarity.