It took 2.30 minutes using a current of 2.00A to plate out all the silver from 0.250L of a solution containing Ag+. What was the original concentration of Ag+ in the solution?
2007-04-11 17:29:02 · 2 answers · asked by ChEMIsTrY ChICkiE 1 in Science & Mathematics ➔ Chemistry
Total charge passed = 2.3 *60 * 2 =276 Coulomb
we have the formula,
C/F = W/E ( F= Faraday= 96500 C)
in case of Ag+, E = equivalent wt = mol. wt
so, mols of Ag+ displaced = C/F = 276/96500 = 2.86xe-3
hence initial concentration of Ag +
in gram ion/ lt = 2.86xe-3/0.25 = 0.01144 gm ion/lt
in normality = 0.01144(N)
2007-04-11 18:09:03 · answer #1 · answered by s0u1 reaver 5 · 0⤊ 0⤋
Total charge = 2.3*60*2 = 276 C
Moles of Ag+ = 276/96500 = 2.86x10^-3
Thus molarity = 2.86x10^-3/(1/4) = 0.0114 M
2007-04-12 00:43:57 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋