Chemistry questions on Volhard Method

Question

A 0.8050 g of arsenic containing pesticide sample was treated with 25.00 ml of 0.0950 mol/L silver nitrat which resulted in the precipitation of Ag₃AsO₄. The excess silver ions were titrated with 28.05 mL of 0.0755 mol/L potassium thiocyanate solution. Calculate the weight percentage of As₂O₃ in the pesticide sample.  

[Molar mass (g/mol): As₂O₃ = 197.841] 

Answer 1

So for this problem the method is as follows: AsO₄^-3 and Ag⁺ precipitate as Ag₃AsO₄. Let's write the equation.

AsO₄^-3_(aq) + 3Ag⁺_(aq) --> Ag₃AsO_4(s)

We then take the remaining solution, which will contain excess silver ions (we can assume all of the AsO4^-3 has been used to make precipitate), and we titrate it with SCN^-, in another precipitation reaction. Let's write that reaction too.

Ag⁺_(aq) + SCN^-_(aq) --> AgSCN_(s)

OK, now we've got the stoichiometry of the problem established, let's get started. We are told that after the arsenic containing pesticide reacts with the Ag⁺, the excess Ag⁺ reacts with 28.05 mL of 0.0755 mol of potassium thiocyanate, KSCN. Let's use that information to figure out how much excess Ag⁺ we had at the end of the first step.

28.05 mL x L x 0.0755 mol x 1 mol Ag⁺ = 0.002118 mol of excess Ag+ after the first step

1000 mL L KSCN 1 mol SCN^-

Great. Now let's talk about the first step, when our arsenic containing pesticide forms solid Ag₃AsO₄. We know how much Ag⁺ we end up with, because we just calculated it, but how much did we start with? We'll need the info they give us about the AgNO₃ solution.

25 mL x 1 L x 0.0950 mol = 0.002375 mol Ag⁺ present at the beginning of first step

1000 mL L

Now we know how much Ag⁺ we start with, and how much we end up with. That's two boxes of an "ICE box" table. We can get the change too by subtracting the initial amount from the equilibrium amount. Let's make the table and see if we can fill in more.

| Reaction | AsO₄^-3 | 3 Ag⁺ | Ag₃AsO₄ |

|-----------------------------|-----------|---------------|--------------|

| Initial (moles) | ? | 0.002375 | 0 |

| Change (moles) | | -0.000257 | |

| Equilibrium (moles) | 0 | 0.002118 | |

The question mark is what we need to know in order to answer this question. We can assume that all of the AsO₄^-3 has reacted, hence the 0 moles at equilibrium. According to the balanced equation, if "x" moles of AsO₄^-3 are reacting, they are reacting with "3x" moles of Ag⁺. So we can take our change in Ag⁺ and divide by three to get the change in AsO₄, and therefore also the initial amount. We can fill in the column for the precipitate as well but there's no need.

| Reaction | AsO₄^-3 | 3 Ag⁺ | Ag₃AsO₄ |

|-----------------------------|---------------|---------------|-------------------|

| Initial (moles) | 0.00008567 | 0.002375 | 0 |

| Change (moles) | -0.00008567 | -0.000257 | +0.00008567 |

| Equilibrium (moles) | 0 | 0.002118 | 0.00008567 |

So we started with 0.00008567 mol of AsO₄^-3 ions, which contains 0.00008567 mol of arsenic. Since the pesticide contained arsenic in the form of As₂O₃, we have 2 mol of arsenic for 1 mol of As₂O₃. Finally we use the molar mass to convert grams to moles.

0.00008567 mol AsO4-3 x 1 mol As x 1 mol As2O3 x 197.841 g = 0.008482 g As₂O₃

1 mol AsO4 2 mol As mol

Finally, we take this mass and plug it into the equation for percent mass.

percent mass = mass of As₂O₃ x 100%

total mass pesticide

percent mass = 0.008482 g x 100% = 1.054%

0.8050 g

And so the percent mass is 1.054%.