Question

7. A 10.00 mL solution of unknown Pb(II) concentration gave a peak voltammetric diffusion limited current of 6.4 microamps. When 10.00 mL of a known 0.01 M Pb(II) standard is added to the original 10.00 mL, the new diffusion limited current is 18.6 microamps. What is the Pb(II) concentration in the original solution?

(A) 0.002 M
(B) 0.005 M
(C) 0.001 M
(D) 0.0011 M
(E) 0.004 M

Answer 1

peak current, ip = C*[Pb(II)]
C = Constant
[Pb(II)] = concentration of Pb(II)

Initially let concentration is x M.
So 6.4E-6A = C*x Equation(1)

After adding 10 ml 0.01M Pb(II) to original 10 ml of x M,
The concentration became, [Pb(II)] = (10*0.01+10*x)/20 = (x+0.01)/2
In second case, current ip = 18.6E-6A
So 18.6E-6 = C*(x+0.01)/2 Equation(2)

Dividing 2 by 1
(18.6/6.4) = (x+0.01)/(2*x)
18.6*2*x = 6.4*(x+0.01)
x(18.6*2-6.4) = 0.01*6.4
x = 0.002M
So answer is (a)