Calculate pCu^2+ at each of the following points in the titration of 25.0 mL of

ID: 928020 • Letter: C

Question

Calculate pCu^2+ at each of the following points in the titration of 25.0 mL of 0.020M MnS04 with 0.0100M EDTA in a solution buffered to pH 11 in a 1.00M concentration on Nh3. Then answer va = 0 mL va = 1.00 mL va = 45.00 mL va = 50.00 mL va = 55.00 mL Why are we using Nh3 in the first place? Calculate [OH-] for this solution explain mathematically why the Cu2+ does not precipitate out of solution Why can't it be used? Which constant: ka, Kb, ksp, kf determines why it cannot be used? Compete this sentence: K of the complex between and must be greater than K of the complex between and must be if indicator cannot be used.

Explanation / Answer

Moles of Cu+2

0.025L x 0.020M = 5x10^-4moles Cu+2

a) V = 0m L

Moles of EDTA added = 0

So concentration of Cu+2 =0.02

pCu+2 = 1.698

b) v =1mL

Moles of EDTA
0.001L x 0.01M EDTA =10^-4moles EDTA

Cu binds EDTA 1:1 ratio
so, 10^-4moles EDTA binds 10^-4moles Cu+2 leaving 4x10^-4moles Cu+2 left over
4 x10^-4moles Cu +2 / 0.026L = 0.0153M
pCu +2 = 1.81

c)

v = 45mL
Moles of EDTA
0.0045L x 0.01M EDTA = 4.5 X 10^-4moles EDTA

Cu binds EDTA 1:1 ratio
so, 4.5 X10^-4moles EDTA binds 4.5 X10^-4moles Cu+2 leaving 0.5 x10^-4moles Cu+2 left over
0.5 x10^-4moles Cu +2 / 0.07 L =7.14 X 10^-4
pCu +2 = 3.14

(d)

v = 50mL
Moles of EDTA
0.005L x 0.01M EDTA = 5 X 10^-4moles EDTA

Cu binds EDTA 1:1 ratio
so, 5 X10^-4moles EDTA binds 5 X10^-4moles Cu+2 leaving no moles Cu+2 left over

e) similarly after addition of 55mL no moles of Cu+2 will left

2) NH3 is used to maintain the pH of solution

3) the pOH of 1M NH3 will be

              NH3 + HOH --> (NH4+) + (OH-)
Initial [ ] 1M                       0M      0M
Change [ ] -x    +x         +x
Equilibrium [ ] 1 -x           x             x
Kb= [NH4+][OH-]/ [NH3]= x^2/ 1-x
Kb= 1.8 x 10^ -5 = x^2/ 1-x

1.8 x 10^ -5 ~ x^2/ 1 x=.0042
x= [OH-] = .0042

so pOH = 2.37

The Ionic product does not exceeds its Ksp

pOH= -log [OH-]= -log(.0013) = 2.87
14- pOH = pH
pH= 11.13

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Calculate pCu^2+ at each of the following points in the titration of 25.0 mL of

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