25 ml of solution of Fe2+ ions was titrated with a solution of the oxi

1.	25 ml of solution of Fe2+ ions was titrated with a solution of the oxidizing agent Cr2O72- ,50 ml of 0.01 M K2Cr2O7 solution was required . what is the molarity of the Fe2+ solution ?
Answer» The equation to the above REACTION is:- Cr2O72- + 6Fe2+ + 14 H+ -> 2 Cr3+ + 6 Fe3+ + 7H20 Given, Concentration of K2Cr2O7 = 0.01 M Volume of K2Cr2O7 = 50 ml Therefore, moles of K2Cr2O7 present = ? M = n/V 0.01M = (n/50)1000 n = (0.01 50 )/1000 = 0.0005 moles 1 mole of K2Cr2O7 -> 6 moles of Fe2+ 0.0005 moles of -> 6 * (0.0005)= 0.003 moles of Fe2+ Therefore, the strength of Fe2+ solution required is =? V= 25ml = 0.025l n= 0.003 moles M=? M= n / V (in LITRES) = 0.003/ 0.025 = 0.12M Ans:- 50ml of 0.01M K2Cr2O7 solution would require 25ml of 0.12M solution of Fe2+.

25 ml of solution of Fe2+ ions was titrated with a solution of the oxidizing agent Cr2O72- ,50 ml of 0.01 M K2Cr2O7 solution was required . what is the molarity of the Fe2+ solution ?

Answer»

The equation to the above REACTION is:-

Cr2O72- + 6Fe2+ + 14 H+ -> 2 Cr3+ + 6 Fe3+ + 7H20

Given, Concentration of K2Cr2O7 = 0.01 M

Volume of K2Cr2O7 = 50 ml

Therefore, moles of K2Cr2O7 present = ?

M = n/V

0.01M = (n/50)*1000

n = (0.01 * 50 )/1000

= 0.0005 moles

1 mole of K2Cr2O7 -> 6 moles of Fe2+

0.0005 moles of -> 6 * (0.0005)= 0.003 moles of Fe2+

Therefore, the strength of Fe2+ solution required is =?

V= 25ml = 0.025l

n= 0.003 moles

M=?

M= n / V (in LITRES)

= 0.003/ 0.025

= 0.12M

Ans:- 50ml of 0.01M K2Cr2O7 solution would require 25ml of 0.12M solution of Fe2+.