Question

In the laboratory a student combines 29.3 mL of a 0.309 M nickel(II) nitrate solution with 22.2 mL of a 0.381 M iron(III) nitrate solution.

What is the final concentration of nitrate anion?

__________ M

Answer 1

nickel(II) nitrate is Ni(NO3)2

iron(III) nitrate is Fe(NO3)3

Concentration of mixture = (n1*C1*V1+ n2*C2*V2) / (V1+V2)

where C1 --> Concentration of 1 component

V1-->volume of 1 component

C2 --> Concentration of other component

V2-->volume of other component

n1 --> number of particle from 1 molecule of 1st component

n2 --> number of particle from 1 molecule of 2nd component

use:

C = (n1*C1*V1+ n2*C2*V2) / (V1+V2)

C = (2*0.309*29.3+3*0.381*22.2)/(29.3+22.2)

C = 0.844 M

Answer: 0.844 M