Question

A 25.0 mL sample containing Ni2+ was analyzed and gave an instrument siganl of 23.6 units (corrected for a blank). When exactly 0.500 mL of 0.0287 Ni(NO3)2 was added to the solution, the signal increased to 37.9 units. Calculate the molar concentration of Ni2+ assuming that the signal was directly proportional to the analyte concentration.

I was told this was a single addition standard addition problem, but I'm not sure how to work it. Thanks!

Answer 1

given

siganl was directly proportional to concentration

so

consider

A = e x C

where e is the proportionality constant

now

initially

23.6 = e x C1

and

finally

37.9 = e x C2

so

23.6 / 37.9 = C1 / C2

C2/C1 = 1.606

now

we know that

moles = molarity x volume (L)

so

initially

moles of Ni+2 = C1 x 25 x 10-3

now

0.5 ml of 0.0287 M Ni+2 is added by Ni(N03)2

so

moles of Ni+2 added = 0.0287 x 0.5 x 10-3

moles of Ni+2 added = 1.435 x 10-5

now

final moles = ( 1.435 x 10-5 ) + ( 25C1 x 10-3)

now

final volume = 25 + 0.5 = 25.05 ml

now

final concentration ( C2) = moles / volume (L)

C2 = ( 1.435 x 10-5 ) + ( 25C1 x 10-3) / 25.05 x 10-3

25.05C2 x 10-3 = 1.435 x 10-5 + ( 25C1 x 10-3)

now

substitute = C2 = 1.606 C1

we get

25.505 x 1.606C1 x 10-3 = 1.435 x 10-5   + ( 25C1 x 10-3)

40.96C1 x 10-3 = 1.435 x 10-5 + ( 25C1 x 10-3)

15.96 x C1 x 10-3 = 1.435 x 10-5

C1 = 9 x 10-4

so

the molar concentration of Ni+2 is 0.0009 M