# EDTA Titration calculation overlook needed

1. Jul 30, 2008

### ssb

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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 $$NiSO_4$$ = $$0.00011 mol * 154.75$$g/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 im getting the hang of this stuff

2. Jul 30, 2008

### Staff: Mentor

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

3. Jul 30, 2008

### ssb

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 $$NiSO_4$$ = $$0.00011 mol * 154.75$$g/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

4. Jul 30, 2008

### Staff: Mentor

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.