Question

Determine the concentration of nitrate ions after 126.0 mL of 0.121 M solution of Cu(NO3)2 and 193.0 mL of 0.215 M solution of Fe(NO3)3 are mixed together. Assume that volumes are additive.

A). 0.887 M

B). 0.486 M

C). 0.336 M

D). 0.168 M

E). 0.178 M

Answer 1

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

n1 = 2 as 1 molecule of Cu(NO3)2 has 2 NO3- ion

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

n2 = 3 as 1 molecule of Fe(NO3)3 has 3 NO3- ion

use:

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

C = (2*0.121*126+3*0.215*193)/(126+193)

C = 0.4858 M

Answer: 0.486 M