Question

a) Salicylate ion + Hexaaquaiton (III) ion react to produce an ion complex, the stoichiometry is 1:1. Using this 0.05g salicyclic acid is diluted to 100 mL using 0.2M iron (III) chloride (FeCl3) solution. What is the concentration of salicyclic acid in this stock solution?

b) using a pipette transfer o.5, 1, 1.5, 2 and 2.5 mL of the stock solution to five 10mL flasks and dilute to mark using 0.02M FeCl3 solution. What is the concentration of standard solutions now prepared?

Answer 1

a)

molarity of solution = moles / volume of solution in liter

mass of salicylic acid = 0.05g

moles of the salicylic acid = mass/molar mass of salicylic acid

the molar mass of salicylic acid = 138.12 g/mol

moles of the salicylic acid = mass/molar mass of salicylic acid

moles of the salicylic acid = 0.05 / 138.12 = 0.000362 mol

since the solution diluted to 100 mL

volume of solution = 100 mL = 0.1 L

molarity of salicylic acid solution = moles / volume of solution in liter

= 0.000362 / 0.1

molarity of salicylic acid solution =  0.00362 M

concentration of salicylic acid in stock solution =  0.00362 M

b)

o.5, 1, 1.5, 2 and 2.5 mL of the stock solution to five 10mL flasks

here we are making a dilution of stock solution, the concentration of solution while making dilution is related by

M1V1 = M2V2

M1 = molarity of the stock solution = 0.00362 M

V1 = volume of stock solution taken

M2 = molarity of the final standard solution made

V2 = volume of the final standard solution made = 10 mL

M1V1 = M2V2

M2 = M1V1 / V2

so for, V1 = volume of stock solution taken = 0.5 mL

M2 = M1V1 / V2

M2 = ( 0.00362 * 0.5 ) / 10

M2 = 0.000181 M

so, concentration of standard solution made by using 0.5 mL stock solution = 0.000181 M  

so for, V1 = volume of stock solution taken = 1 mL

M2 = M1V1 / V2

M2 = ( 0.00362 * 1 ) / 10

M2 = 0.000362 M

so, concentration of standard solution made by using 0.5 mL stock solution = 0.000362 M  

so for, V1 = volume of stock solution taken = 1.5 mL

M2 = M1V1 / V2

M2 = ( 0.00362 * 1.5 ) / 10

M2 = 0.000543

so, concentration of standard solution made by using 0.5 mL stock solution = 0.000543​​​​​​​ M  

so for, V1 = volume of stock solution taken = 2 mL

M2 = M1V1 / V2

M2 = ( 0.00362 * 2 ) / 10

M2 = 0.000724

so, concentration of standard solution made by using 0.5 mL stock solution = 0.000724 M  

so for, V1 = volume of stock solution taken = 2.5 mL

M2 = M1V1 / V2

M2 = ( 0.00362 * 2.5 ) / 10

M2 = 0.000905 M

so, concentration of standard solution made by using 0.5 mL stock solution = 0.000905 M