ALTERNATIVE TO PRACTICAL A
A is a solution of KMnO\(_4\) containing 0.020 moldm\(^{-3}\). B is a solution of Fe\(^{2+}\) obtained by dissolving 3.8 g of iron granules in 250 cm\(^3\) of solution.
(a)Put A in the burette. Pipette 20.0 cm\(^3\) or 25 cm\(^3\) of B into a conical flask and add 10 cm\(^3\) of H\(_2\)SO\(_4\). Titrate it with A.
Repeat the titration to obtain concordant titre values. Tabulate your result and calculate the average volume of A used.
MnO\(_4\)\(^-\)\(_{(aq)}\) + 5Fe\(^{2+}\)\(_{(aq)}\) + 8H\(^+\)\(_{(aq)}\) → Mn\(^{2+}\)\(_{(aq)}\) + Fe\(^{3+}\)\(_{(aq)}\) + 4H\(_2\)O\(_{(l)}\)
[ KMnO\(_4\) = 158.0, Fe = 56.0 ]
(b) From your results and information provided, calculate the:
(i) concentration of B in moldm\(^{-3}\)
(ii)concentration of B in gdm\(^{-3}\)
(iii) mass of Fe\(^{2+}\) in 250 cm\(^3\) of B.
(iv) percentage of Fe\(^{2+}\) in the granules.
(a) Volume of pipette used (Volume of B used) = 25.0 cm\(^3\)
| Burette Readings (cm\(^3\)) | Rough | 1st titre | 2nd titre |
| Final burette readings | 18.20 | 18.50 | 38.40 |
| Initial burette readings | 0.00 | 0.40 | 20.20 |
| Volume of A used (KMnO\(_4\)) | 18.20 | 18.10 | 18.20 |
Volume of A used = \(\frac{Rough + 1st titre + 2nd titre}{3}\)
= \(\frac{18.20 + 18.10 + 18.20}{3}\)
= 18.166
V\(_A\) = 18.17 cm\(^3\)
(i) Concentration of B in moldm\(^{-3}\)
From the balanced equation : MnO\(_4\)\(^-\)\(_{(aq)}\) + 5Fe\(^{2+}\)\(_{(aq)}\) + 8H\(^+\)\(_{(aq)}\) → Mn\(^{2+}\)\(_{(aq)}\) + Fe\(^{3+}\)\(_{(aq)}\) + 4H\(_2\)O\(_{(l)}\)
C\(_A\) = 0.020 moldm\(^{-3}\) N\(_A\) = 1
V\(_A\) = 18.17cm\(^3\) N\(_B\) = 5
C\(_B\) = ?
V\(_B\) = 25.0cm\(^3\)
Using \(\frac{C_A V_A}{C_B V_B}\) = \(\frac{N_A}{N_B}\)
C\(_B\) = \(\frac{C_A V_A N_B}{N_A V_B}\)
C\(_B\) = \(\frac{ 0.02 X 18.17 X 5}{25 X 1}\)
C\(_B\) = 0.07268
C\(_B\) = 0.070 moldm\(^{-3}\)
(ii) Concentration of B in gdm\(^{-3}\)
If 3.8 g of Fe granules was dissolved in 250 cm\(^3\) of B
Xg of Fe granules will be dissolved in 1000 cm\(^3\) of B
X = \(\frac{ 3.8 X 1000}{250}\) = 15.2 gdm\(^{-3}\)
Concentration of B in gdm\(^{-3}\) will be 15.2 gdm\(^{-3}\)
(iii) Mass of Fe\(^{2+}\) in 250cm\(^3\) of B.
To get the mass, we'll use the formula n = \(\frac{mass}{Molar mass}\)
However, since do not know the number of moles, n , hence we'll need to obtain n, using n = \(\frac{CV}{1000}\) since we have C and V
So, n = \(\frac{{0.070}\times{250}}{1000}\) = 0.025 mole
Now, we can apply n = \(\frac{mass}{Molar mass}\) to obtain the mass, given that the Molar mass of Fe was given to be 56
n = \(\frac{mass}{Molar mass}\)
m = n x M = 0.018 x 56 = 1.02 g
Therefore, 1.02 g of Fe\(^{2+}\) in 250 cm\(^3\) of B.
(iv) Percentage of Fe\(^{2+}\) in the granules = \(\frac{Mass of Fe^{2+}}{Total mass of Fe granules}\times 100\)
= \(\frac{1.02}{3.8}\times 100\)
= 26.89%
= 26.9%
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