EDTA Titration calculation overlook needed

[ssb]

Homework Statement

25.00 Ml sample of a solution of Al_2(SO4)3 and NiSO4 was diluted to 500.0 mL and the Al^3+ and Ni^2+ in a 25.00 mL aliquat were complexed by addition of 40.00 mL of 0.01175 M EDTA in a pH 4.8 buffer.

The excess EDTA was then back titrated with 10.07 mL of 0.00993 M Cu^2+. an excess of F- was then added to the solution to displace the EDTA that was bouned to the Al^3+. This liberated EDTA consumed 26.30 mL of 0.00993 M Cu^2+.

Calculate the mg/ml of Al2(SO4)3 (342.17 g/mol) and NiSO4 (154.75 g/mol) in the sample

Homework Equations

The Attempt at a Solution

Calculate moles EDTA = 0.040 L * 0.01175 M = 0.0047 mol EDTA

Calculate moles of Excess EDTA 0.01007 L * 0.00993 M = .000099995 moles excess EDTA

Calculate Moles EDTA reacted with Al^{3+} and Ni^{2+} = 0.0047 - .000099995 = 0.00037 mol total EDTA used

Moles Al^{3+} comsumed = 0.02630 L * 0.00993 M = 0.00026 moles AL^{3+} consumed

moles Ni^{2+} consumed = 0.00037 mol EDTA - 0.00026 mol Mol Al^{3+} = 0.00011 mol Ni^{2+}

Calculate mass Al_2(SO_4)_3 = 0.00026 mol * 342.17 g/m = 0.08896 grams = 8.896 mg

Calculate mass NiSO_4 = 0.00011 mol * 154.75g/m = 0.01702 g = 1.702 mg

Mg/Ml Al_2(SO_4)_3 = 8.896/25.00 = 0.3558 mg/ml

Mg/Ml NiSO_4 = 1.702/25.00 = 0.0681 mg/ml

Do my calculations look correct? sig figs? If so I think I am getting the hang of this stuff

 

[Borek]

Not bad, but not perfect either.

0.00026 moles of Al³⁺, not Al₂(SO₄).

You took 25/500 of the original sample.

0.017 g is not 1.7 mg

 

[ssb]

Not bad, but not perfect either.

0.00026 moles of Al³⁺, not Al₂(SO₄).

You took 25/500 of the original sample.

0.017 g is not 1.7 mg

Borek
--
equation balancer and stoichiometry calculator

Calculate moles EDTA = 0.040 L * 0.01175 M = 0.0047 mol EDTA

Calculate moles of Excess EDTA 0.01007 L * 0.00993 M = .000099995 moles excess EDTA

Calculate Moles EDTA reacted with Al^{3+} and Ni^{2+} = 0.0047 - .000099995 = 0.00037 mol total EDTA used

Moles Al^{3+} comsumed = 0.02630 L * 0.00993 M = 0.00026 moles AL^{3+} consumed

moles Ni^{2+} consumed = 0.00037 mol EDTA - 0.00026 mol Mol Al^{3+} = 0.00011 mol Ni^{2+}

CORRECTION step:
Convert moles Al_2(SO_4)_3 into moles Al^+
Al_2 ----> 2Al^+ + 2e^- therefore there is one mole of Al_2(SO_4)_3 for every 2 moles of Al^+

Calculate mass Al_2(SO_4)_3 = (0.00026* mol *Al^+) (\frac {1 mol Al_2(SO_4)_3} {2 mol Al^+} )* (342.17 g/m) = 0.04448 grams = 44.48 mg

Calculate mass NiSO_4 = 0.00011 mol * 154.75g/m = 0.01702 g = 17.02 mg

If there were 44.48 mg Al_2(SO_4)_3 in 25.00 mL then there are 889.6 mg Al_2(SO_4)_3 in 500 mL

And if there were 17.02 mg NiSO_4 in 25.00 mL then there are 34.04 mg NiSO_4 in 500 mL

Mg/Ml Al_2(SO_4)_3 = 889.6 mg / 500.0 ml = 1.779 mg/ml

Mg/Ml NiSO_4 = 34.04/500.0 = 0.6808 mg/ml

 

[Borek]

What is Al⁺? Where did you get it from? And what for? And what is Al₂? And this redox reaction... Sorry, but it looks like you are trying to solve titration questions having no idea about the most basic ideas.

17.02 in 25 doesn't mean 34.04 in 500.

Note: I am not checking everything, I am not calculating everything by myself, I am just writing about things obvious enough to be spotted at the first sight.

Too much spoonfeeding.