Show all work for points. Sulfide ion was determined by indirect titration with
ID: 788748 • Letter: S
Question
Show all work for points.
Sulfide ion was determined by indirect titration with EDTA. To a solution containing 25.00 mL of 0.043 32 M Cu(CIO4)2 plus 15 mL of 1 M acetate buffer (pH 4.5) were added 25.00 mL of unknown sulfide solution with vigorous stirring. The CuS precipitate was filtered and washed with hot water. Then ammonia was added to the filtrate (which contained excess Cu2+) until the blue color of Cu(NH3) was observed. Titration of the filtrate with 0.039 27 M EDTA required 12.11 mL to reach the murexide end point. Calculate the molarity of sulfide in the unknownExplanation / Answer
PART 1
Titration of Ni2+ and Zn2+ with EDTA and the excess unreacted EDTA was back-titrated with Mg2+. (1) Determine the number of mmoles of EDTA that reacted since this is equal to the total number of mmoles of Ni2+ and Zn2+.
Then,
[(25.0 mL)(0.0452 M EDTA) - (12.4 mL)(0.0123 M Mg2+)] = 0.97748 mmol EDTA = 0.97748 total mmol of Ni2+ and Zn2+
Next, we were told that 2,3-dimercapto-1-propanol was used to displace EDTA from zinc. So we now have liberated EDTA that reacted with another volume of Mg2+. We determine the number of mmol of Mg2+ used since this is equal to the mmol of liberated EDTA which in turn, is the mmol of Zn2+.
(29.2 mL)(0.0123M) = 0.35916 mmol Zn2+
Right away, we can compute for the molarity of Zn2+ by dividing its mmol by the mL of the solution.
0.35916 mmol Zn2+/50.00 mL solution = 0.0071832 M Zn2+
therefore, molarity of Zn2+ is 0.00718 M
To get the mmol Ni2+, we subtract the total mmol of Ni2+ and Zn2+ by the mmol of Zn2+:
0.97748 mmol - 0.35916 mmol = 0.61832 mmol Ni2+
Divide the mmol Ni2+ by mL of solution to get M Ni2+:
0.61832 mmol Ni2+ / 50.00 mL solution = 0.0123664 M Ni2+
therefore, molarity of Ni2+ is 0.0124 M
PART 2
first, we have to calculate molar concentration of Cu2+
[Cu2+]filtrate = (0.03927 M * 12.11 mL)/25 mL = 0.019022
[Cu2+]filtrate = 0.01902 M = 1.902 * 10^-2 M
and then
molar concentration of the sulfide in unknown sample [S^-2] = [Cu(ClO4)2] - [Cu2+]filtrate
= 0.04332 M(given) - 0.01902M
= 0.02430M
= 2.430 * 10^-2 M