A 50.0 mL sample containing Cd2 and Mn2 was treated with 58.7 mL of 0.0500 M EDTA. Titration of the excess unreacted EDTA required 17.7 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 29.1 mL of 0.0300 M Ca2 . What were the molarities of Cd2 and Mn2 in the original solution?
0999
To determine the molarities of Cd2+ and Mn2+ in the original solution, we need to analyze the given information and the reactions involved.
Let's break down the problem step by step:
1. The initial sample contains Cd2+ and Mn2+ ions.
2. The sample was treated with 58.7 mL of 0.0500 M EDTA (ethylenediaminetetraacetic acid), which reacted with the metal ions Cd2+ and Mn2+. EDTA can form a complex with these metal ions.
The reactions can be written as:
Cd2+ + EDTA ⇌ Cd(EDTA)
Mn2+ + EDTA ⇌ Mn(EDTA)
Since we don't know the concentrations of Cd2+ and Mn2+ in the original solution, we'll assign them as "x" and "y" moles respectively in 50.0 mL of the solution.
So, [Cd2+] = x/0.0500 L (since 50.0 mL = 0.0500 L) and [Mn2+] = y/0.0500 L.
3. Any excess unreacted EDTA is titrated with 17.7 mL of 0.0300 M Ca2+ (calcium ion). This means that the Ca2+ ions react with the remaining EDTA.
The reaction can be written as:
Ca2+ + EDTA ⇌ Ca(EDTA)
4. The Cd2+ ions are displaced from the EDTA complex by the addition of an excess CN− (cyanide) ions. This reaction forms a soluble Cd(CN)4^2- complex.
The reaction can be written as:
Cd(EDTA) + 4CN− ⇌ Cd(CN)4^2- + EDTA
5. The newly freed EDTA is titrated with 29.1 mL of 0.0300 M Ca2+ ions. This means that the remaining EDTA reacts with the added Ca2+ ions.
The reaction can be written as:
Ca2+ + EDTA ⇌ Ca(EDTA)
Now that we have all the reactions, we can determine the number of moles of EDTA consumed in each step using stoichiometry, and then calculate the moles of Cd2+ and Mn2+ in the original solution.
Let's start with the calculations:
1. Calculate the moles of EDTA used in the first titration:
moles EDTA used in first titration = volume of 0.0500 M EDTA * concentration of EDTA
= 58.7 mL * 0.0500 mol/L
= 2.93 mmol
2. Calculate the moles of Ca2+ used in the first titration:
moles Ca2+ used in first titration = volume of Ca2+ titrant * concentration of Ca2+
= 17.7 mL * 0.0300 mol/L
= 0.531 mmol
3. Calculate the moles of EDTA used in the second titration:
moles EDTA used in second titration = volume of 0.0300 M Ca2+ * concentration of Ca2+
= 29.1 mL * 0.0300 mol/L
= 0.873 mmol
4. Calculate the moles of Ca2+ used in the second titration:
moles Ca2+ used in second titration = volume of Ca2+ titrant * concentration of Ca2+
= 29.1 mL * 0.0300 mol/L
= 0.873 mmol
Now, let's set up the equations based on the reactions and solve for the unknowns:
For the first titration:
moles EDTA used = moles Ca2+ used
2.93 mmol = 0.531 mmol + moles of Ca2+ used to titrate Cd(EDTA)
moles of Ca2+ used to titrate Cd(EDTA) = 2.93 mmol - 0.531 mmol
= 2.399 mmol
Since 1 mole of Cd(EDTA) reacts with 4 moles of Ca2+ ions, we have:
moles of Cd(EDTA) = (1/4) * moles of Ca2+ used to titrate Cd(EDTA)
= (1/4) * 2.399 mmol
= 0.5997 mmol
Using the molar mass of Cd(EDTA) (308.35 g/mol), we can calculate its molarity in the original solution:
[Cd(EDTA)] = moles of Cd(EDTA) / volume of solution in liters
= 0.5997 mmol / 0.0500 L
= 11.99 mmol/L
= 11.99 mM
Now we can calculate the molar concentration of Cd2+ in the original solution:
[Cd2+] = [Cd(EDTA)]
Therefore, the molarity of Cd2+ in the original solution is 11.99 mM.
To find the molarity of Mn2+ in the original solution, we need to consider that Mn2+ reacts with EDTA in a 1:1 ratio. Therefore, the moles of Mn2+ in the original solution are equal to the moles of EDTA used in the first titration, which is 2.93 mmol.
Using the volume of the solution (50.0 mL = 0.0500 L), we can calculate the molar concentration of Mn2+ in the original solution:
[Mn2+] = moles of Mn2+ / volume of solution in liters
= 2.93 mmol / 0.0500 L
= 58.6 mmol/L
= 58.6 mM
Therefore, the molarity of Mn2+ in the original solution is 58.6 mM.