The concentration of Sn2+ in a solution is determined by titration with a 0.1650
ID: 492938 • Letter: T
Question
The concentration of Sn2+ in a solution is determined by titration with a 0.1650 M permanganate solution. The balanced net ionic equation for the reaction is:
2MnO4-(aq) + 5Sn2+(aq) + 16H3O+(aq) 2Mn2+(aq) + 5Sn4+(aq) + 24H2O(l)
(a) If 15.75 mL of the 0.1650 M permanganate solution are needed to react completely with 30.00 mL of the Sn2+ solution, what is the concentration of the Sn2+ solution? _________M
(b) Which of the two solutions was in the buret during the titration?
(c) Suppose at the end of the titration, the solution containing the Mn2+ ion is transferred to a volumetric flask and diluted to 400. mL. What is the concentration of Mn2+ in the diluted solution? ________M
Explanation / Answer
2MnO4-(aq) + 5Sn2+(aq) + 16H3O+(aq) -------------------->2Mn2+(aq) + 5Sn4+(aq) + 24H2O(l)
2 moles 5 moles
MnO4^- Sn^+2
M1 = 0.165M M2 =
V1 = 15.75ml V2 = 30ml
n1 = 2 n2 = 5
M1V1/n1 = M2V2/n2
M2 = M1V1n2/n1V2
= 0.165*15.75*5/2*30 = 0.2165M
Conc of Sn^+2 = 0.2165M
b. KMnO4 solution taken in the buret.
c.
Initial Fianl
M1 = 0.165M M2 =
V1 = 15.75ml V2 = 400ml
M1V1 = M2V2
M2 = M1V1/V2
= 0.167*15.75/400 = 0.0066M
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