Week 7: Discussion
David Fabian
05/14/2014
Warm‐up: EDTA Titration
A 0.550 L solution contains 0.466 g of MgSO4 (FW = 120.37 g/mol). You take a 49.8 mL aliquot of MgSO4 and titrate with EDTA. 42.0 mL of EDTA is required to reach the end point of the titration. In a second titration, you use the same EDTA solution to titrate into a CaCO3 solution. What mass, in milligrams, of CaCO3 (FW = 100.09 g/mol) will react with 20.0 mL of EDTA? Assume that EDTA reacts with both MgSO4 and CaCO3 in 1:1 ratios.
HINT: First determine [EDTA] to reach endpoint
A 0.550 L solution contains 0.466 g of MgSO4 (FW = 120.37 g/mol). You take a 49.8 mL aliquot of MgSO4 and titrate with an EDTA solution. 42.0 mL of EDTA is required to reach the end point of the titration. In a second titration, you use the same EDTA solution to titrate into a CaCO3 solution. What mass, in milligrams, of CaCO3 (FW = 100.09 g/mol) will react with 20.0 mL of EDTA? Assume that EDTA reacts with both MgSO4 and CaCO3 in 1:1 ratios.
Checkpoint:
0.466 g
49.8 mL aliquot
0.000351 mol Mg 2 0.351 mmol Mg 2 0.351 mmol EDTA
120.37 g / mol
550 mL total
0.351 mmol EDTA
0.00835 M EDTA
42 mL EDTA
A 0.550 L solution contains 0.466 g of MgSO4 (FW = 120.37 g/mol). You take a 49.8 mL aliquot of MgSO4 and titrate with an EDTA solution. 42.0 mL of EDTA is required to reach the end point of the titration. In a second titration, you use the same EDTA solution to titrate into a CaCO3 solution. What mass, in milligrams, of CaCO3 (FW = 100.09 g/mol) will react with 20.0 mL of EDTA? Assume that EDTA reacts with both MgSO4 and CaCO3 in 1:1 ratios.
0.466 g
49.8 mL aliquot
0.000351 mol Mg 2 0.351 mmol Mg 2 0.351 mmol EDTA
120.37 g / mol
550 mL total
0.351 mmol EDTA
0.00835 M EDTA
42 mL EDTA
0.00835
mmol
EDTA 20.0 mL 0.167 mmol
mL
0.167 mmol 100.09
mg
16.7 mg
mmol
Mg‐EDTA Complexation
Consider a Na2MgY (Y4‐ = EDTA) solution with a total concentration of 1.50 x 10‐3 M and total EDTA concentration of 0.0100 M. The alpha fraction for Y4‐ is 0.236 (because some EDTA is protonated). For the MgY2‐ complex, log Kf = 8.69. Assume Na+ ions are spectator ions, and EDTA reacts with Mg2+ in a 1:1 ratio
Calculate pMg (= ‐log[Mg2+]). HINT: K ’f = α * Kf = [Y4‐]/[EDTAT] * {[MgY2‐]/([Mg2+][Y4‐])}
Mg‐EDTA Complexation
Consider a Na2MgY (Y4‐ = EDTA) solution with a total concentration of 1.50 x 10‐3 M and total EDTA concentration of 0.0100 M. The alpha fraction for Y4‐ is 0.236 (because some EDTA is protonated). For the MgY2‐ complex, log Kf = 8.69. Assume Na+ ions are spectator ions, and EDTA reacts with Mg2+ in a 1:1 ratio.
Calculate pMg (= ‐log[Mg2+]). Mg2+ + Y4‐ MgY2‐
We have some fraction of Y4‐, so conditional formation constant, K ’f is:
K ’f = α * Kf = [MgY2‐]/([Mg2+][EDTAT])
Rearrange to solve for [Mg2+]:
[Mg2+] = [MgY2‐]/(α*Kf*[EDTAT])
= 0.0015/(0.236*(108.69)*0.01)
[Mg2+] = 1.30*10‐9
pMg = ‐log(1.30*10‐9) = 8.89
Mg‐EDTA Complexation – Part 2
Consider a Na2MgY (Y4‐ = EDTA) solution with a total concentration of 1.50 x 10‐3 M. The alpha fraction for Y4‐ is really small, 1.8 * 10‐11 (because we’re at a pH where almost all EDTA is protonated). For the MgY2‐ complex, log Kf = 8.69. Assume Na+ ions are spectator ions, and EDTA reacts with Mg2+ in a 1:1 ratio.
Calculate pMg (= ‐log[Mg2+]). Mg2+ + Y4‐ MgY2‐
Now, MgY2‐ is the only source of Mg2+ and EDTA. Thus [EDTAT] = [Mg2+]
K ’f = αKf = [MgY2‐]/([Mg2+][EDTAT])
αKf = [MgY2‐]/([Mg2+]2)
[Mg2+]2 = [MgY2‐]/(αKf) [Mg2+] = sqrt([MgY2‐]/(αKf))
= sqrt{0.0015/((1.8 * 10‐11)*(108.69))}
[Mg2+] = 0.412; pMg = 0.385
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