Question

10.00 ml of a solution of an analyte gives a signal of 30 mV in an instrumental analysis. after the addition of 2e-5 mole standard, the signal rises to 75 mV. what is the concentration of the analyte in the original solution? assuming the total volume doesn't change.

Answer 1

E=Eo-RT/nF lnQ (nerst equation)

Where,

Eo=standard reduction potential

R=gas constant

F=faraday’s constant

N=number of electrons in redox reaction

Q=reactant quotient=activity of product/activity of reactant

Or, E=Eo-(0.05916 /n ) lnQ (nerst’s equation)

Given Q=a(1 1 1

10.00 ml of a solution of an analyte gives a signal of 30 mV in an instrumental analysis. after the addition of 2e-5 mole standard, the signal rises to 75 mV. what is the concentration of the analyte in the original solution? assuming the total volume doesn't change.

Concentration of analyte=[analyte]=moles of analyte/10 ml=m/0.01 L

30mV=Eo-(0.05916 /n ) ln [m/0.01 L]….(1)

Also 75 mV=Eo-(0.05916 /n ) ln [m+(2*10^-5 moles)/0.01 L]……………..(2)

Equation (2)-(1) gives,

45*10^-3=0.05916/n ln [m/0.01 L]- ln [m+(2*10^-5 moles)/0.01 ]

0.045 =0.05916/n ln [m/ m+(2*10^-5 moles)]

Or,0.761=ln [m/ m+(2*10^-5 moles)]                                 for [n=1]

Or,exp(0.761)= [m/ m+(2*10^-5 moles)]

2.14= m/ m+(2*10^-5 moles)

2.14m+4.28*10^-5=m

1.14m=4.28*10^-5

m=3.75*10^-5

[analyte]=moles of analyte/10 ml=m/0.01 L=3.75*10^-5/0.01=3.75 *10^-3 mol/L