The edta stock solution is 5.503 x 10^-5 M. Please show all work. Will rate. Thi
ID: 1073213 • Letter: T
Question
The edta stock solution is 5.503 x 10^-5 M. Please show all work. Will rate. This is for an analytical chemistry class. So Determination (20 Points: 10 points titration, 10 points error/statistics) Sulfate was determined through an indirect titration. A 10.00 mL aliquot of tap water was treated with excess Ba(NO32 to form a precipitate at a pH of 1.00. This precipitate was filtered and washed before being boiled with 125.00 mL of the standardized EDTA at a pH of 10.00. Excess EDTA was titrated with Mg2 at a pH of 10.00 using Eriochrome Black T indicator. SO4 Titration Trial V Mg (mL) 21.31 22.8 20.32 23.26 18.83 23.19 22.69 22.87 22.55 23.2Explanation / Answer
1 mole of EDTA reacts with 1 mole of Ba2+
excess EDTA reacts with Mg2+
concentration of sulfate = total EDTA - EDTA reacted with Mg2+
Trial 1
Total EDTA = 5.503 x 10^-5 M x 125 ml
= 6.9 x 10^-3 mmol
lets say molarity of Mg2+ solution is same as EDTA solution molarity, then,
Excess unreacted EDTA = 5.503 x 10^-5 M x 21.31 ml
= 1.2 x 10^-3 mmol
moles of sulfate present = EDTA reacted
= 6.9 x 10^-3 - 1.2 x 10^-3
= 5.7 x 10^-3 mmol
mass of sulfate present = 5.7 x 10^-3 mmol x 96.06 g/mol
= 0.55 mg in 10 ml tap water
[please note : exact molarity of Mg2+ is needed to be fed above to get actual concentration of sulfate in 10 ml tap water sample].
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