Calculate pMn 2+ at each of the following points in the titration of 26.65 mL of
ID: 529480 • Letter: C
Question
Calculate pMn2+ at each of the following points in the titration of 26.65 mL of 0.0212 M EDTA with 0.0106 M MnCl2. The EDTA solution is buffered at a pH of 10.00. The fraction of free EDTA in the Y4- (ay4) can be form found here. The formation constant for the Mn -EDTA complex is given by log Kf 13.89
Link for free EDTA: https://sites.google.com/site/chempendix/fraction-of-free-edta-as-y4
A) 5.330 mL
B) 21.32 mL
C) 42.64 mL
D) 52.23 mL
E) 53.19 mL
F) 53.30 mL
G)53.41 mL
H) 58.63 mL
I) 63.96 mL
Explanation / Answer
First we need to calculate the conditional constant using the value of alpha (a = 0.30) and Kf:
Kf* = Kf*alpha(pH 10)
Kf* = 0.30*1013.89
logKf* = 13.367
Next, we calculate how many moles of EDTA we have in the flask we are titrating:
26.65mL EDTA * (0.0212 mol EDTA / 1000mL EDTA) = 0.00056498 mol EDTA
Now we need to calculate how many moles of Mn2+ are added in each of the following volumes (the total volume is calculated just adding the 26.65mL of EDTA and the volume of Mn2+ solution added):
A) 5.330mL (total volume = 31.98mL)
5.330mL Mn2+ * (0.0106 mol Mn2+ / 1000mL Mn2+ ) = 0.000056498 mol Mn2+
The reaction between EDTA and Mn2+ is 1:1.
We sketch the following table to make things easier:
Our concentrations are:
(EDTA) = (0.000508482mol)/(0.03198L) = 0.0159M
(Mn2+) = 0
(MnEDTA) = (0.000056498mol)/(0.03198L) = 0.00176M
However, we have to consider the back reaction (MnEDTA decomposing into Mn2+ and EDTA), so our concentrations change:
(EDTA) = (0.000508482mol)/(0.03198L) = 0.0159M + x
(Mn2+) = x
(MnEDTA) = (0.000056498mol)/(0.03198L) = 0.00176M - x
We calculate X using the equilibrium expression:
Kf * = (Mn-EDTA) / (Mn2+)(EDTA)
Kf * = (0.00176 - x) / (x)(0.0159 + x)
Kf* = (0.00176 - x) / (0.0159x + x2)
Kf* (0.0159x + x2) = 0.00176 - x
Kf*x2 + 0.0159Kf*x + x - 0.00176 = 0
1013.367x2 + 1011.568x - 0.00176 = 0
Solve for x
(Mn2+) = x = 4.76 x 10-15 M
We can finally calculate pMn using that concentration:
pMn = -log (4.76 x 10-15)
pMn = 14.32
B) 21.32mL (total volume = 47.97mL)
21.32mL Mn2+ * (0.0106 mol Mn2+ / 1000mL Mn2+ ) = 0.000225992 mol Mn2+
We sketch the following table to make things easier:
Our concentrations are:
(EDTA) = (0.000338988mol)/(0.04797L) = 0.00706M
(Mn2+) = 0
(MnEDTA) = (0.000225992mol)/(0.04797L) = 0.00471M
However, we have to consider the back reaction (MnEDTA decomposing into Mn2+ and EDTA), so our concentrations change:
(EDTA) = 0.00706M + x
(Mn2+) = x
(MnEDTA) = 0.00471M - x
We calculate X using the equilibrium expression:
Kf * = (Mn-EDTA) / (Mn2+)(EDTA)
Kf * = (0.00471 - x) / (x)(0.00706 + x)
Kf* = (0.00471 - x) / (0.00706x + x2)
Kf* (0.00706x + x2) = 0.00471 - x
Kf*x2 + 0.00706Kf*x + x - 0.00471 = 0
1013.367x2 + 1011.216x - 0.00471 = 0
Solve for x
(Mn2+) = x = 2.86 x 10-14 M
We can finally calculate pMn using that concentration:
pMn = -log (2.86 x 10-14)
pMn = 13.54
C) 42.64mL (total volume = 69.29mL)
42.64mL Mn2+ * (0.0106 mol Mn2+ / 1000mL Mn2+ ) = 0.000451984 mol Mn2+
We sketch the following table to make things easier:
Our concentrations are:
(EDTA) = (0.000112996mol)/(0.06929L) = 0.00163M
(Mn2+) = 0
(MnEDTA) = (0.000451984mol)/(0.06929L) = 0.00652M
However, we have to consider the back reaction (MnEDTA decomposing into Mn2+ and EDTA), so our concentrations change:
(EDTA) = 0.00163M + x
(Mn2+) = x
(MnEDTA) = 0.00652M - x
We calculate X using the equilibrium expression:
Kf * = (Mn-EDTA) / (Mn2+)(EDTA)
Kf * = (0.00652 - x) / (x)(0.00163 + x)
Kf* = (0.00652 - x) / (0.00163x + x2)
Kf* (0.00163x + x2) = 0.00652 - x
Kf*x2 + 0.00163Kf*x + x - 0.00652 = 0
1013.367x2 + 1010.579x - 0.00652 = 0
Solve for x
(Mn2+) = x = 1.72 x 10-13 M
We can finally calculate pMn using that concentration:
pMn = -log (1.72 x 10-13)
pMn = 12.76
D) 52.23mL (total volume = 78.88mL)
52.23mL Mn2+ * (0.0106 mol Mn2+ / 1000mL Mn2+ ) = 0.000553638 mol Mn2+
We sketch the following table to make things easier:
Our concentrations are:
(EDTA) = (0.000011342mol)/(0.07888L) = 0.000144M
(Mn2+) = 0
(MnEDTA) = (0.000553638mol)/(0.07888L) = 0.00702M
However, we have to consider the back reaction (MnEDTA decomposing into Mn2+ and EDTA), so our concentrations change:
(EDTA) = 0.000144M + x
(Mn2+) = x
(MnEDTA) = 0.00702M - x
We calculate X using the equilibrium expression:
Kf * = (Mn-EDTA) / (Mn2+)(EDTA)
Kf * = (0.00702 - x) / (x)(0.000144 + x)
Kf* = (0.00702 - x) / (0.000144x + x2)
Kf* (0.000144x + x2) = 0.00702 - x
Kf*x2 + 0.000144Kf*x + x - 0.00702 = 0
1013.367x2 + 109.525x - 0.00702 = 0
Solve for x
(Mn2+) = x = 2.09 x 10-12 M
We can finally calculate pMn using that concentration:
pMn = -log (2.09 x 10-12)
pMn = 11.68
EDTA Mn2+ Mn-EDTA I 0.00056498 0.000056498 C 0.000056498 0.000056498 0.000056498 E 0.000508482 0 0.000056498Related Questions
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