Volhard titration was used to find out the mass% of i-in a 0.6532-g sample. a 50.00-ml of 0.05539 m silver nitrate solution was added to the sample and the precipitate was allowed to form. back titration of the remaining silver required 33.12 ml of 0.05233 mpotassium thiocyanate. determinethe mass% of i-(atomic weight 126.9 amu) in the sample.

The mass% of iodide ions in the sample is 20.13%.

Explanation:

Total Moles of silver nitrate added = n

Molarity of the silver nitrate solution = 0.05539 M

Volume of the silver nitrate = 50.00mL =  0.050 L

..[1]

Moles of potassium thiocyanate = n'

Molarity of the potassium thiocyanate  solution = 0.05233 M

Volume of the potassium thiocyanate = 33.12 mL=  0.03312 L

According to reaction-1, 1 mole of potassium thiocyanate reacts with 1 mol of silver nitrate. Then 0.0017332 mole of potassium thiocyanate reacts with 0.0017332 mol of silver nitrate.

Moles of silver nitrate which had reacted with iodide ions = N

n = n' + N

N = n - n' = 0.0027695 mol - 0.0017332 = 0.0010363 mol

..[2]

According to reaction-2, 1 mole of iodide reacts with 1 mol of silver nitrate. Then 0.0010363 mole of silver nitrate reacts with :

of silver nitrate.

Mass of 0.0010363 mole of iodide ions =

0.0010363 mol × 126.9 g/mol = 0.1315 g

The mass% of iodide ions in the sample:

Mass of sample = 0.6532 g

i'm pretty sure its b i might be wrong though!