Calculate pCd2 at each of the following points in the titration of 40.00 mL of 0.0040 M Cd2 with 0.0040 M EDTA in the presence of the auxiliary complexing agent NH3. The solution is buffered at a pH of 11.00. Take the NH3 concentration to be a constant 0.100 M. The formation constant for the Cd2 -EDTA complex is given by log Kf = 16.5.
a) 0 mL
b) 3.00mL
c) 35.00 mL
d) 40.00mL
e) 45mL
To calculate pCd2 at each point in the titration, we need to determine the amount of Cd2+ that has reacted with EDTA at each point. Here's how to calculate pCd2 at each of the given points in the titration:
a) 0 mL:
At this point, no EDTA has been added yet, so there is no reaction between Cd2+ and EDTA. Therefore, pCd2 is equal to the initial concentration of Cd2+, which is 0.0040 M.
b) 3.00 mL:
To calculate pCd2 at this point, we need to determine the amount of Cd2+ remaining after reacting with EDTA. Since EDTA is added in a 1:1 stoichiometric ratio to Cd2+, we can calculate the number of moles of Cd2+ that have reacted as follows:
Number of moles of Cd2+ reacted = concentration of EDTA × volume of EDTA added
= 0.0040 M × 0.0030 L (since 3.00 mL is equal to 0.0030 L)
= 0.012 mol
Now, we need to subtract this amount from the initial concentration of Cd2+:
Remaining moles of Cd2+ = initial moles of Cd2+ - moles of Cd2+ reacted
= (0.0040 M × 0.0400 L) - 0.012 mol (since there are 40.00 mL initially)
= 0.148 mol
Finally, we can calculate pCd2 at this point by dividing the remaining moles of Cd2+ by the new total volume (40.00 mL + 3.00 mL) and taking the negative logarithm:
pCd2 = -log(base 10) (remaining moles of Cd2+ / total volume)
= -log(base 10) (0.148 mol / 0.043 L)
≈ -log(base 10) (3.442 M)
≈ -0.537
c) 35.00 mL:
Following the same process as in part b, calculate the moles of Cd2+ reacted by multiplying the concentration of EDTA by the volume of EDTA added:
Moles of Cd2+ reacted = 0.0040 M × 0.0350 L (since 35.00 mL is equal to 0.0350 L)
= 0.140 mol
Subtract this amount from the initial moles of Cd2+:
Remaining moles of Cd2+ = initial moles of Cd2+ - moles of Cd2+ reacted
= (0.0040 M × 0.0400 L) - 0.140 mol
= 0.000 mol (all Cd2+ has reacted)
Since all Cd2+ has reacted, the pCd2 value is essentially zero (pCd2 = -log(0) = infinity).
d) 40.00 mL:
At this point, all Cd2+ has reacted with EDTA, so there is no Cd2+ left in solution, resulting in a pCd2 value of infinity.
e) 45 mL:
Since 45.00 mL is beyond the initial volume of the solution (40.00 mL), this point falls outside the range of the given problem and cannot be determined.
In summary:
a) pCd2 = -log(0.0040 M) = 2.397
b) pCd2 ≈ -0.537
c) pCd2 = ∞ (all Cd2+ has reacted)
d) pCd2 = ∞ (all Cd2+ has reacted)
e) Cannot be determined.