Question

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.

Answer 1

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

Step-by-step 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

`Moles=Molarity* Volume (L)`

`n= 0.05539 M* 0.050 L=0.0027695 mol`

`AgNO_3+KSCN\rightarrow AgSCN+KNO_3`..[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

`n'= 0.05539 M* 0.03312 L=0.0017332 mol`

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

`I^-+AgNO_3\rightarrow AgI+NO_(3)^-`..[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 :

`(1)/(1)0.0010363 mol= 0.0010363 mol` 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

`=\frac{\text{Mass of iodide ions}}{\text{mass of sample}}* 100`

`=(0.1315 g)/(0.6532g)* 100=20.13\%`