Question

What volume of 0.146 M Mg(NO3)2 should be added to 255ml of 0.102M KNO3 to produce a concentration of solution 0.278 M NO3- ions? Assume volumes are addictive

Answer 1

Answer:

Explanation:

 We need to consider the 0.416M Mg(NO3)2 solution for the standpoint of just the nitrate ion, in which case the concentration is twice the 0.416 value. So, [NO3^-] in the Mg(NO3)2 solution is 0.832 M.

M1V1 + M2V2 = M3V3

(0.832 mol/L) (x) + (0.102 mol/L) (255 mL) = (0.278 mol/L) (255 + x)

0.832x + 26.01 = 70.89 + 0.278x

0.554x = 44.88

x = 81.01083 mL

Answer 2

Answer: 81.01083 mL of 0.146 M Mg(NO3)2 should be added to 255ml of 0.102M KNO3 to produce a concentration of solution 0.278 M NO3- ions.

Explanation:

To view the 0.416M Mg(NO3)2 solution from the perspective of the nitrate ion alone, where the concentration is twice that of 0.416, is necessary.

In the Mg(NO3)2 solution, [NO3-] thus equals 0.832 M.

The formula to be applied is

=M1V1 + M2V2 = M3V3

=(0.832 mol/L) (x) + (0.102 mol/L) (255 mL) = (0.278 mol/L) (255 + x)

=0.832x + 26.01 = 70.89 + 0.278x

=0.554x = 44.88

x = 81.01083 mL

So, the correct volume to be added is 81.01083 mL.

To know more about the concentration of the solution from the given link

https://brainly.com/question/17206790

https://brainly.com/question/26255204