Chemistry question on Volhard method

Question

The Volhard method was used for the following analysis:

A 2.0412 g sample that contains only sodium chloride and barium iodide was dissolved and diluted to a volume of 250.00 mL volumetric flask. A 25.00 mL aliquot of the sample was treated with 39.02 mL of a 0.09876 M AgNO3 solution and the excess Ag+ was back-titrated with 13.98 mL of a 0.1045 M SCN−.

[Molar masses (g mol −1): NaCl = 58.443; BaI2 = 391.15]

1.1.1 Calculate the weight percentages of NaCl and BaI2 in the sample.

1.1.2 Which indicator will be used to detect the endpoint in a Volhard titration?

1.1.3 Use reaction equations to illustrate how the endpoint is detected in Volhard titration.

1.1.4 The AgCl solid is removed by filtration before the back-titration of the excess silver nitrate. Explain why this is necessary. You may use reaction equations to aid your explanation.

Answer 1

After we prepare our solution of NaCl and BaI₂, we've got aqueous Cl^- and I^- ions. Then we add Ag⁺ to this solution in the form of AgNO₃, forming insoluble AgCl and AgI (you can look at their K_sp's to confirm they are insoluble, they are very small). So we have these two reactions:

Ag⁺ + I^- → AgI_(s)

Ag⁺ + Cl^- → AgCl_(s)

Let's go ahead and use 'X' to generalize, X will stand for either halogen, I or Cl.

Ag⁺ + X^- → AgX_(s)

Then we filter out the solid, and the remaining solution contains excess Ag⁺. This solution is titrated with SCN^-, forming another precipitate, insoluble AgSCN. We can figure out how much Ag⁺ was left in solution after the first reaction, by how much SCN^- it takes to titrate that solution. So let's figure out how many moles of SCN- we add during the titration, since that will be how many moles of Ag⁺ were present at the beginning of the titration.

13.98 mL x 1L/1000mL x (0.1045 mol)/(L SCN^-) x (1 mol Ag⁺)/(1 mol SCN^-) = 0.00146091 mol of Ag⁺

This is how much Ag⁺ we have at the beginning of the titration, so it's also the amount of excess Ag⁺ left over after the first precipitation reaction with the halogens. We can also figure out how much Ag⁺ was added at the beginning of the first precipitation reaction, using the info given about the AgNO₃ solution.

39.02 mL x 1L/1000mL x (0.09876 mol AgNO₃)/(L) x (1 mol Ag⁺)/(1 mol AgNO₃) = 0.0038536 mol of Ag⁺

We now know how much Ag⁺ we start with, how much is left at equilibrium, and therefore also the amount that reacted with the halide ions. Let's fill in an "ICE" table for our generalized reaction between Ag⁺ and X^-.

Ag+ + X- --> AgX(s)

0.0038536

-0.0023927

0.0014610

Since both reactants have coefficients of one, we know the change in X^- is the same as the change in Ag⁺. We can also assume that no X^- ions are left at equilibrium. So let's fill in the column for X^-.

Ag+ + X- --> AgX(s)

0.0038536 0.0023927

-0.0023927 -0.0023927

0.0014610 0

So now we know that there 0.0023927 mol of X^- there to react with the Ag⁺. But remember that we didn't use the entire sample for this reaction, we only used 25 mL of the solution prepared from that sample. Knowing the moles and volume, we can get the concentration of that solution.

[X^-] = 0.0023927 mol / 0.025 L = 0.095708 M

We also know the entire solution was 250 mL, allowing us to calculate the number of moles of X^- to make that solution.

n(X^-) = 0.095708 mol/L x 0.25 L = 0.023927 mol

This is the number of moles of Cl^- and I^- combined. We were also given the total mass of the sample. So we can write two equations in two unknowns. We need to be careful though to note that moles of I^- will be double the moles of BaI₂, because there are two I^- ions per formula unit of BaI₂.

0.023927 = n(Cl^-) + n(I^-)

n(Cl^-) = n(NaCl), n(I^-) = 2 x n(BaI₂), make substitutions

0.023927 = n(NaCl) + 2 x n(BaI₂), rearrange

n(NaCl) = 0.023927 - 2 x n(BaI₂)

2.0412 = m(NaCl) + m(BaI₂)

mass = molar mass x moles

2.0412 = 58.443 x n(NaCl) + 391.15 x n(BaI₂), substitute in n(NaCl)

2.0412 = 58.443 x (0.023927 - 2 x n(BaI₂)) + 391.15 x n(BaI₂), do a bunch of algebra

2.01412 = 1.39837 - 116.886 x n(BaI₂) + 391.15 x n(BaI₂)

0.642834 = 274.26 x n(BaI₂)

n(BaI₂) = 0.002344 mol, now go back and get n(NaCl)

n(NaCl) = 0.023927 - 2 x 0.002344

n(NaCl) = 0.019239 mol

We know how many moles of each we have! The first part of the question asks for weight percentages, so we need to multiply by the molar mass to get masses.

m(BaI₂) = 0.91680 g

m(NaCl) = 1.1244 g

Sanity check before we calculate the mass percent... do your masses of BaI₂ and NaCl add to equal 2.0412 g? Do your amounts in moles of Cl^- and I^- add to equal 0.023927? They should! I'll let you do that check for yourself.

%M(NaCl) = 1.1244 g / 2.0412 g x 100% = 55.086% NaCl

%M(BaI₂) = 0.916798 g / 2.0412 g x 100% = 44.915% BaI₂

So that's 1.1.1. Now let's talk about 1.1.2. The indicator that will be useful here will be one that changes color when we run out of Ag⁺ to react with, and the SCN^- concentration begins to build up. For the Volhard method you would usually use the ferric ion Fe³⁺. This ion forms a complex with SCN^-, FeSCN⁺². The aqueous Fe3+ will be light yellow, but the complex formed with SCN^- is dark red. (pretend the arrow below is an equilibrium arrow rather than a double headed arrow that is usually used for resonance structures)

Fe³⁺ + SCN^- ↔ FeSCN²⁺

As for 1.1.4, AgCl must be filtered out to avoid the common ion effect. Why AgCl but not AgI? Well you can't really filter out one without also filtering the other, but AgI shouldn't cause the same common ion effect problem. Why? Because of their K_sp's.

AgSCN(s) → Ag+ + SCN^- K_sp = 1.03 x 10^-12

AgCl(s) → Ag+ + Cl^- K_sp = 1.6 x 10^-10

AgI(s) → Ag+ + I^- K_sp = 8.3 x 10^-17

So while AgSCN is more soluble than AgI, it is less soluble than AgCl. This means than the SCN^-, by reacting with Ag⁺, could cause some of the AgCl to dissolve back into Ag⁺ and Cl^-. The result would be whatever amount of excess Ag⁺ we calculate based on the titration is artificially higher than the true amount, and the rest of our calculations would be thrown off too.