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.
Explanation
| Rough Tree | 1st Titre | 2rd Titre | 3rd Titre | |
| Final reading cm\(^3\) | 15.60 | 31.10 | 45.50 | 15.60 |
| Initial reading cm\(^3\) | 0.00 | 15.60 | 31.10 | 0.00 |
| Volume of the acid cm\(^3\) | 15.50 | 15.50 | 15.40 | 15.60 |
(a) Average Titre = \(\frac{15.50 + 15.40 + 15.60}{3}\) cm\(^3\)
= \(\frac{46.50}{3}\)cm\(^3\)
= 15.50 cm\(^3\)
(b)(i) Concentration of B in moldm\(^{-3}\)
250cm\(^3\) of B contains 480g of FeCl\(_2\)
100 cm\(^{-3}\) of B will contain \(\frac{4.80}{250}\) x 1000g
= 19.2 gdm\(^{-3}\)
Concentration of B in moldm\(^{-3}\) ?
= \(\frac{\text{conc. in gdm}^{-3}}{\text{molar mass}} = \frac{19.2 gdm^{-3}}{127 gmol^{-1}}\)
= 0.151 mol dm\(^3\)
Concentration of A in mol dm\(^{-3}\) = ?
\(\frac{C_AV_A}{C_BV_B} = \frac{n_A}{n_B}\)
C\(_A\) = Molar concentration in moldm\(^{-3}\) of KMnO\(_4\) = ?
V\(_A\) = Volume of KMnO\(_4\) used = 15.50cm\(^3\)
n\(_A\) = 1
C\(_B\) = Molar concentration of FeCl\(_{2(aq)}\) = 0.151 moldm\(^{-3}\)
V\(_B\) = Volume of FeCl\(_{2(aq)}\) = 25.00 cm\(^3\)
= \(\frac{C_A \times 15.50 cm^3}{0.151 mold dm^{-3} \times 25.00 cm^3}\) = \(\frac{1}{5}\)
C\(_A = \frac{0.151 \times 1 \times 25.00}{5 times 15.50} = \frac{3.775}{77.50}\)
= 0.0487 (3sf)
C\(_A\) = 0.0487 moldm\(^{-3}\)
(ii) Number of moles of Fe\(^{2+}\) in the volume of B pipetted? 1000 cm\(^3\) of B contains 0.151 moles of Fe\(^{2+}\) 25cm of B will contain \(\frac{0.151}{1000} \times 25\) moles = 0.0003775 mole of Fe\(^{2+}\)