A 1.000-mL aliquot of a solution containing Cu2+ and Ni2* is treated with 25.00

Question

A 1.000-mL aliquot of a solution containing Cu2+ and Ni2* is treated with 25.00 mL of a 0.04346 M EDTA solution. The solution is then back titrated with 0.02272 M Zn2 solution at a pH of 5. A volume of 19.22 mL 2+ Ni2 solution is fed through an ion-exchange column that retains Ni2+. The Cu2+ that passed through the column is treated with 25.00 mL 0.04346 M EDTA. This solution required 21.01 mL of 0.02272 M Zn2* for back titration. The Ni2+ extracted from the column was treated witn 25.00 mL of 0.04346 M EDTA How many milliters of 0.02272 M Zn2 is required for the back titration of the Ni2 solution? Number mL

Explanation / Answer

Step 1: Aliquote of Ni2+ and Cu2+ treated with 25 mL of 0.04346 M solution of EDTA (EDTA will react with Cu2+ and Ni2+ completely and the remaining EDTA will be there in the solution)

This EDTA is titrated against Zn+ solution having 0.02272 M strength with 19.22mL

That means the remaining EDTA needed 19.22 mL of 0.0272 M solution of Zn2+ for complete reacation.

Now let calculate the strength volume of EDTA left in the solution

V1, Volume of Zn2+ = 19.22 mL

N1, Strength of Zn2+ = 0.02272

V2, Volume of EDTA = ?

N2, Strength of EDTA = 0.04346

Now use N1V1 = N2V2

So V2 = (19.22 X 0.02272)/0.04346 = 10.0478 mL

It means 10.0478 mL of EDTA is remained in the solution after the reaction with Cu2+ and Ni 2+

Hence the volume of EDTA needed to react with Cu2+ and Ni2+ = 25-10.0478 = 14.9522 mL

Now let's calculate the strength of Cu2+ and Ni2+ using these values:

V1, Volume of EDTA = 14.9522

N1, Strength of EDTA = 0.04346

V2, Vol. of Cu2+ and Ni2+ = 1.0000mL

N2, Strength of Cu2+ and Ni2+ = ?

According to the equation, N2 = 0.6489 M

In second experiment:

Voulume of Cu2+ and Ni2+ = 2.0000mL

Cu2+ (alon after ellution) treated with 25 mL of 0.04346M solution of EDTA.

After complete reaction with Cu2+, EDTA remains there and it is treated with 21.01 mL of 0.02272 M solution of Ni2+,

First let's find out the volume of EDTA remained in the solution

V1, Volume of Zn2+ used = 21.01 mL

N1 Strength of Zn2+ used = 0.02272 M

V2, Volume of EDTA =?

Strength of EDTA = 0.04346 M

By the previous equation, V2 = 10.9836 mL

It means 10 9836 mL of EDTA was remained in the solution to react with Zn2+.

Hence the volume of EDTA required to react with Cu2+ solution = 25 - 10.9836 mL = 14.0164 mL

Now let's find out the normality of this Cu2+ solution.

V1, Volume of EDTA = 14.0146

N1, Strength of EDTA = 0.04346

V2, Volume of Cu 2+ = 2.000 mL

N2, Strength of Cu2+ =?

Use the euuation, N2 = 0.03055 M

In second case the volume of aliquote was 2.0000 mL and it need 14.0164 mL EDTA to react only with Cu2+

Hence the volume of EDTA to react only with Cu2+ 1.000 mL of aliquote = 14.0164/2 = 7.0082 mL

In first case the volume of EDTA used for both Cu2+ and Ni2+ = 14. 9522

In which 7.0082 mL react with Cu2+ and the remaining EDTA react with Ni2+

SO the volume of EDTA reacted with Ni2+(in fiurst case) = 7.944 mL

Let's calculate the strength of Ni2+ from this value:

V1, Volume of EDTA used = 7.944

N1, Strength of EDTA = 0.04346

V2, Volume of Ni2+ = 1.000 mL

N2, Strength of Ni2+ =?

Use the euuation, N2 = 0.3452 M

Now let's apply this strength to second experiment.

V1, Volume of Ni2+ = 2.00 mL (second case aliquote)

N1, Strength of NI2+ = 0.3452

V2, Volume of Zn2+ = ?

N2, Strength of Zn2+ =0.02272

Using the equation, V2 = 30.3849 mL

Hence the volume 0.0722 M Zn2+ needed to react with ellueted Ni2+ = 30.3849 mL

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