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Consider the titration 50.0 ml of 5.00 x 103 m Cd2+ with 0.0100 m EDTA at a ph of 10. Calculate the pcd when 35.00 ml of 0.010 m EDTA has been added. Kcsy 2.9x 10, 40.35 a) 9.24 b) 9.09 c) 12.30 d) 15.61

Accepted Answer

The correct answer is A) 9.24. Here's how to calculate the pCd:
1. Calculate moles of Cd2+ and EDTA:

Moles of Cd2+ = (50.0 mL)(5.00 x 10^-3 M) = 0.250 mmol
Moles of EDTA = (35.00 mL)(0.0100 M) = 0.350 mmol
2. Determine the moles of Cd2+ remaining:

Since EDTA is in excess, the reaction goes to completion. The moles of Cd2+ remaining = 0.250 mmol - 0.350 mmol = -0.100 mmol.
Since we can't have negative moles, this means all the Cd2+ has reacted, and we have an excess of EDTA.
3. Calculate the concentration of free EDTA:

Excess EDTA = 0.350 mmol - 0.250 mmol = 0.100 mmol
[EDTA] = 0.100 mmol / (50.0 mL + 35.00 mL) = 1.25 x 10^-3 M
4. Use the conditional formation constant (Kf') to calculate the concentration of free Cd2+:

Kf' = Kf
αY4- = 2.9 x 10^16
(0.35) = 1.015 x 10^16 (αY4- is the fraction of EDTA in the Y4- form at pH 10, approximately 0.35)
[Cd2+] = [EDTA] / Kf' = (1.25 x 10^-3 M) / (1.015 x 10^16) = 1.23 x 10^-19 M
5. Calculate pCd:

pCd = -log[Cd2+] = -log(1.23 x 10^-19) = 18.91
Important Note: The given answer choices are incorrect. The calculated pCd is 18.91, not one of the options. This may indicate an error in the problem statement or answer choices.


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