The solutions at the two Pb electrodes of a concentration cell were prepared as
ID: 503852 • Letter: T
Question
The solutions at the two Pb electrodes of a concentration cell were prepared as follows: Cell A: A mixture of 1.00 mL of 0.0500 M Pb(NO3) 2 with 4.00 mL of 0.0500 M KX (the soluble potassium salt of an unspecified monovalent ion X Some PbX2(s) precipitates. Cell B: 5.00 mL of 0.0500 M Pb(NO3)2. The cell potential was measured to be 0.06700 V at 25 OC. By use of the Nernst equation, determine the concentration (M) of Pb n the solution of cell A 2 pts Submit Answer Tries 0/99 In Cell A, how many moles of X have reacted with 2+ 2 pts Submit Answer Tries 0/99 What is the concentration (M) of X in the solution of cell A. 2 pts Submit Answer Tries 0/99Explanation / Answer
E = -0.059/n log [anodic concentration/cathodic concentration]
or, 0.067 = 0.059/2 log [Pb^2+ in cell A/ Pb^2+ in cell B]
or, [Pb^2+ in cell A] = 0.00027 M
---------------------------
Pb^2+ + 2X- ==> PbX2
moles of Pb^2+ present intially = 0.001L * 0.05 M = 5*10^-5 moles
moles of Pb^2+ present after adding X- = 0.00027 M
moles of Pb^2+ precipitated = 5*10^-5 moles - ( 0.00027 M * 0.005 L) = 4.865*10^5 moles
To precipitate Pb^2+ twice the amount X^- is required . So, moles of I- required = 2* 4.865*10^5 moles
= 9.73*10^-5 moles
Moles of X- present in the solution = (0.004 L *0.05 M) - (9.73*10^-5 moles) = 1.027 x 10^-4 moles
Ksp = [Pb++] [X-]^2 = (0.00027 M)( 0.02054 M)^2 = 5.5 x 10^-6