A 50.0 mL sample containing Cd2 and Mn2 was treated with 49.1 mL of 0.0700 M EDTA. Titration of the excess unreacted EDTA required 17.5 mL of 0.0300 M Ca2 . The Cd2 was displaced from EDTA by the addition of an excess of CN–. Titration of the newly freed EDTA required 10.9 mL of 0.0300 M Ca2 . What were the molarities of Cd2 and Mn2 in the original solution?
By taking the total amount of EDTA (3.44 mmol), and subtracting the consumed portion (0.525 mmol), I got 2.91 mmol as my Cd2 + Mn2 mmol.
How do I take this, and calculate Cd2 by itself, and the Mn2 from that?
Total = 3.44-0.525 = ?
Mn^2+ alone = 10.9 x 0.03 = ? mmol.
I took that answer (0.327 mmol) and divided it by 50ml to get my Molarity. But this does not seem correct. What am I doing wrong?
Probably not anything. I didn't read the problem right. The 10.9 mL x 0.03 is "the newly freed EDTA from the CN^- complex with Cd^2+) so this is the Cd and the difference is Mn.
That is, Cd alone is the 10.9 x 0.03 mmol and convert that to M as usual. Mn is the difference between total and Cd.
To calculate the molarities of Cd2+ and Mn2+ in the original solution, you need to use the information provided about the titration and the reaction stoichiometry. Here are the steps to follow:
1. Calculate the moles of EDTA used:
- Volume of EDTA used = 49.1 mL
- Concentration of EDTA = 0.0700 M
- Moles of EDTA = volume (in L) × concentration = 0.0491 L × 0.0700 mol/L = 0.003437 mol
2. Calculate the moles of Ca2+ used in the titration of excess unreacted EDTA:
- Volume of Ca2+ used = 17.5 mL
- Concentration of Ca2+ = 0.0300 M
- Moles of Ca2+ = volume (in L) × concentration = 0.0175 L × 0.0300 mol/L = 0.000525 mol
3. Calculate the moles of EDTA that reacted with Cd2+ and Mn2+ ions:
- Moles of EDTA reacted = Moles of EDTA used - Moles of Ca2+ used = 0.003437 mol - 0.000525 mol = 0.002912 mol
4. Since 1 mole of Cd2+ reacts with 2 moles of EDTA, you can calculate the moles of Cd2+ present in the original solution:
- Moles of Cd2+ = 0.002912 mol / 2 = 0.001456 mol
5. Subtract the moles of Cd2+ from the total moles of Cd2+ + Mn2+ to obtain the moles of Mn2+:
- Moles of Mn2+ = Total moles - Moles of Cd2+ = 0.002912 mol - 0.001456 mol = 0.001456 mol
6. Finally, to calculate the molarities, divide the moles by the volume of the original solution (in L):
- Volume of the original solution = 50.0 mL = 0.0500 L
- Molarity of Cd2+ = Moles of Cd2+ / Volume of original solution = 0.001456 mol / 0.0500 L = 0.0291 M
- Molarity of Mn2+ = Moles of Mn2+ / Volume of original solution = 0.001456 mol / 0.0500 L = 0.0291 M
Therefore, the molarities of Cd2+ and Mn2+ in the original solution are both 0.0291 M.