Question

Analyte in an unknown gave a signal of 25.0 mV. When 1.00 mL of 0.065 M standard addition reagent was added to

10.00 mL of the unknown sample, the signal increased to 35.0 mV. What is the analyte concentration of the original

unknown?

Answer 1

Initial volume,V = 10.00 ml

Total volume after addition,V' = 10.00 + 1.00 = 11.00 ml

let initial concentration = X and Final concentration = X'

X' = X (V'/V) = X (11.00/10.00)

so X' = X (1.10)

Similarly

let known concentration of standard before addition = S = 0.065M and

known concentration of standard after addition= S'

S' = S (V'/V) = 0.065 (11.00/10.00) =0.065 (1.10) M

Given initial signal , I = 25.0 mV and final signal, I' = 35.0 mV

The formula is

X = 0.7143(1.1X +0.0715) = 0.7857X + 0.05107

X = 0.05107/0.2143 = 0.2382 M

On solving, we get initial concentration of the original unknown = 0.2382 M