Question

1. A 0.238 g sample contained only NaCl and KBr. It was dissolved in water and required

48.40 mL of 0.048 M AgNO3 for complete titration of both halides [giving AgCl(s) and
AgBr(s)]. Calculate the weight percent of Br in the solid sample.
2. A 5.000 g sample of pesticide was decomposed with metallic sodium in alcohol and the
liberated chloride ion was precipitated as silver chloride. Express the result of this
analysis in terms of percent DDT (C14H9Cl5) based on the recovery of 0.1606 g of
precipitate.
3. Find the pCl in a 20 mL of a 0.10 M Cl- solution after addition of 0, 10, 20, and 30 mL of
0.10 M AgNO3 . Ksp = 1.0x10-10

Answer 1

The question requires using the principles of precipitation titration and stoichiometry to find the mass percent of chloride in a sample by reacting it with a known concentration and volume of AgNO3 solution.

Step-by-step explanation:

The question you've asked involves multiple concepts in analytical chemistry, in particular the precipitation titration of halides and calculating weight percent compositions based on chemical reactions. In these problems, silver nitrate (AgNO3) is used to titrate chloride ions, forming insoluble silver chloride (AgCl), among other precipitation reactions.

For the example where a 0.200 g sample of chloride was titrated with AgNO3, we would first calculate the moles of AgNO3 used in the titration:
Moles of AgNO3 = Molarity × Volume (L) = 0.1 M × 0.028 L = 0.0028 moles.

Since each mole of AgNO3 reacts with one mole of chloride ions, the mass of chloride ions (Cl−) can be determined and used to calculate the mass percent of chloride in the sample. We use the molar mass of Cl− to convert moles to grams and then find the mass percent by comparing it to the initial mass of the sample.

Answer 2

The weight percent of Br in the solid sample is approximately 8.42%.

Step-by-step explanation:

Calculate the moles of AgNO3 used:

Moles AgNO3 = Volume × Concentration

Moles AgNO3 = 48.40 mL × 0.048 M

Moles AgNO3 = 0.0023192 g

Separate reactions for AgCl and AgBr:

AgNO3 + NaCl → AgCl + NaNO3 (1 mol AgNO3 reacts with 1 mol Cl-)

AgNO3 + KBr → AgBr + KNO3 (1 mol AgNO3 reacts with 1 mol Br-)

Calculate the moles of Cl- and Br- in the sample:

Moles Cl- = Moles AgNO3 used in reaction 1

Moles Cl- = 0.0023192 g / 143.32 g/mol (molecular weight of AgCl)

Moles Cl- = 0.00001618 mol

Moles Br- = Moles AgNO3 used - Moles Cl- (assuming all remaining AgNO3 reacted with Br-)

Moles Br- = 0.0023192 g - 0.00001618 mol

Moles Br- = 0.00230302 mol

Calculate the weight of Br- and its percentage:

Weight Br- = Moles Br- × Atomic weight of Br

Weight Br- = 0.00230302 mol × 79.904 g/mol

Weight Br- = 0.018411 g

Weight percent Br = (Weight Br- / Weight sample) × 100%

Weight percent Br = (0.018411 g / 0.238 g) × 100%

Weight percent Br ≈ 8.42%

Therefore, the weight percent of Br in the solid sample is approximately 8.42%.