A 10.0 ml diluted chloride sample required 44.89 ml of 0.01982 M AgNO3 to reach the Fajans end point. how many moles of Cl- ions were present in the sample? What was the concentration of chloride in the diluted solution?
mol = M x L = ?
M Cl^- = mols Cl^-/L sample
To find the number of moles of Cl- ions present in the sample, we can use the stoichiometry of the reaction between silver nitrate (AgNO3) and chloride ions (Cl-). The balanced equation for the reaction is:
AgNO3 + Cl- -> AgCl + NO3-
From the equation, we can see that one mole of AgNO3 reacts with one mole of Cl- to form one mole of AgCl.
First, let's calculate the number of moles of AgNO3 used in the reaction. We are given the volume and concentration of AgNO3. The equation to calculate the number of moles (n) is:
n = concentration x volume
n = 0.01982 M x 44.89 ml = 0.01982 M x 0.04489 L = 0.000888 mol
Since the stoichiometry of the reaction is 1:1 between AgNO3 and Cl-, the number of moles of Cl- ions present in the sample is also 0.000888 mol.
To find the concentration of chloride (Cl-) in the diluted solution, we need to know the volume of the diluted solution.
Let's assume the diluted solution is obtained by diluting a certain volume of the original solution to a total volume of V ml.
The original solution and diluted solution have the same number of moles of chloride ions. Therefore, we can set up the equation:
M1 x V1 = M2 x V2
where M1 represents the original concentration of chloride, V1 is the original volume, M2 is the diluted concentration, and V2 is the diluted volume.
Rearranging the equation gives us:
M2 = (M1 x V1) / V2
We are given the V1 (10.0 ml) and M2 is what we want to find, so we need to find the V2.
We know that the diluted solution required 44.89 ml of 0.01982 M AgNO3 to reach the endpoint. Therefore, the total volume of the diluted solution is 44.89 ml.
Substituting the values into the equation:
M2 = (M1 x V1) / V2
M2 = (M1 x 10.0 ml) / 44.89 ml
Now, we can calculate the concentration of chloride (Cl-) in the diluted solution. Let's assume the original concentration of chloride (M1) is x.
M2 = (x x 10.0 ml) / 44.89 ml
Simplifying the equation gives us the concentration of chloride in the diluted solution (M2).
Please substitute the value of M1 (original concentration of chloride) to get the final answer.