Question

In the laboratory a student combines 48.1 mL of a 0.405 M sodium iodide solution with 24.0 mL of a 0.527 M sodium nitrate solution. What is the final concentration of sodium cation ?

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

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

we have below equation to be used:

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

C = (1*0.405*48.1+1*0.527*24)/(48.1+24)

C = 0.4456 M

Answer: 0.446 M