TEST OF PRACTICAL KNOWLEDGE QUESTION
All your burette readings (initials and final) as well as the size of your pipette must he recorded but no account of experimental procedure is required. All calculations must be done in your booklet.
A solution of potassium tetraoxomanganate( VII). B is a solution of iron(II)chloride containing 4.80g of the salt in 250 cm\(^{3}\) of solution.
(a) Put A into the burette. Pipette 20.0cm\(^3\) or 2.50.0 of B into a conical flask, add 20.0 cm\(^3\) of H\(_2\)SO4\(_{(aq)}\) and titrate with A. Repeat the titration to obtain concordant titre values. Tabulate your results and calculate the average volume of A used. The equation of the reaction is: MnO\(_{4(aq)}\) + 5Fe\(^{3+}_{ (aq)}\) + H\(_2\))
(b) From your results and the information provided, calculate the;
(i) concentration of B moldm\(^{-3}\):
(ii) concentration of A in moldm\(^{-3}\)
(iii) number of moles of Fe\(^{2+}\) in the volume of B pipetted. [FeCl\(_2\) = 127 gmol\(^{-1}\)] Credit will be given for strict adherence to the instruction, for observations precisely recorded and for accurate inferences. AlI tests, Observations and inferences must be clearly entered in the booklet in ink at the time they are made.
Ionic equation for the reaction:
\[ \text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \to \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O} \]
(a) Burette readings and average titre of A
| Titration | Rough | 1st | 2nd | 3rd |
|---|
| Final reading (cm3) | 15.50 | 15.50 | 30.90 | 15.60 |
| Initial reading (cm3) | 0.00 | 0.00 | 15.50 | 0.00 |
| Volume of A used (cm3) | 15.50 | 15.50 | 15.40 | 15.60 |
Average titre \( = \dfrac{15.50 + 15.40 + 15.60}{3} = \dfrac{46.50}{3} = 15.50\ \text{cm}^3 \).
(b)(i) Concentration of B in mol dm-3
250 cm3 of B contains 4.80 g of FeCl2, so in g dm-3:
\[ \frac{4.80}{250} \times 1000 = 19.2\ \text{g dm}^{-3} \]
\[ C_B = \frac{19.2}{127} = 0.151\ \text{mol dm}^{-3} \]
(b)(ii) Concentration of A in mol dm-3
From the equation the mole ratio \( \dfrac{n_A}{n_B} = \dfrac{1}{5} \), and \( \dfrac{C_A V_A}{C_B V_B} = \dfrac{n_A}{n_B} \), with \( V_A = 15.50\ \text{cm}^3 \), \( V_B = 25.0\ \text{cm}^3 \):
\[ C_A = \frac{C_B \times V_B \times 1}{V_A \times 5} = \frac{0.151 \times 25.0}{15.50 \times 5} = \frac{3.775}{77.50} = 0.0487\ \text{mol dm}^{-3} \]
(b)(iii) Number of moles of Fe2+ in the 25.0 cm3 pipetted
\[ n(\text{Fe}^{2+}) = 0.151 \times \frac{25.0}{1000} = 3.78 \times 10^{-3}\ \text{mol} \]
Ionic equation for the reaction:
\[ \text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \to \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O} \]
(a) Burette readings and average titre of A
| Titration | Rough | 1st | 2nd | 3rd |
|---|
| Final reading (cm3) | 15.50 | 15.50 | 30.90 | 15.60 |
| Initial reading (cm3) | 0.00 | 0.00 | 15.50 | 0.00 |
| Volume of A used (cm3) | 15.50 | 15.50 | 15.40 | 15.60 |
Average titre \( = \dfrac{15.50 + 15.40 + 15.60}{3} = \dfrac{46.50}{3} = 15.50\ \text{cm}^3 \).
(b)(i) Concentration of B in mol dm-3
250 cm3 of B contains 4.80 g of FeCl2, so in g dm-3:
\[ \frac{4.80}{250} \times 1000 = 19.2\ \text{g dm}^{-3} \]
\[ C_B = \frac{19.2}{127} = 0.151\ \text{mol dm}^{-3} \]
(b)(ii) Concentration of A in mol dm-3
From the equation the mole ratio \( \dfrac{n_A}{n_B} = \dfrac{1}{5} \), and \( \dfrac{C_A V_A}{C_B V_B} = \dfrac{n_A}{n_B} \), with \( V_A = 15.50\ \text{cm}^3 \), \( V_B = 25.0\ \text{cm}^3 \):
\[ C_A = \frac{C_B \times V_B \times 1}{V_A \times 5} = \frac{0.151 \times 25.0}{15.50 \times 5} = \frac{3.775}{77.50} = 0.0487\ \text{mol dm}^{-3} \]
(b)(iii) Number of moles of Fe2+ in the 25.0 cm3 pipetted
\[ n(\text{Fe}^{2+}) = 0.151 \times \frac{25.0}{1000} = 3.78 \times 10^{-3}\ \text{mol} \]
