All your burette readings (initials and final) as well as the size of your pipette must be recorded but no account of experimental procedure is required. All calculations must be done in your answer booklet
A solution of 0.050 moldm\(^3\) H\(_2\)C\(_2\)O\(_4\) (ethanedioic acid). B is a solution of KMnO\(_4\), (potassium tetraoxomanganate (VII), of unknown concentration.
(a) Put B into the burette. Pipette 20.0 cm\(^3\) or 25.0 cm\(^3\) of A into a Conical flask and add about 10.0 cm\(^3\) of dilute H\(_2\)SO\(_4\), Heat the mixture to about 40°C - 50°C and titrate it while still hot with B. Repeat the titration to obtain consistent titre values. Tabulate your results and calculate the average volume of B used. The equation of reaction is;
2MnO\(_{4(aq)}^-\) + 5C\(_2\)O\(_{4(aq)}^{2-}\) + 16H\(^+_{(aq)}\) \(\to\) 2MnH\(^{2+}_{(aq)}\) + 8H\(_2\)O\(_{(l)}\) + 10CO\(_{2(g)}\)
(b) From your results and the information provided, calculate the:
(i) concentration of MnO\(_2^-\) in B in moldm\(^{-1}\)
(ii) concentration of KMnO\(_4^-\) in B in gdm\(^{-3}\)
(iii) volume of CO\(_2\) evolved at s.t.p when 25.0 cm\(^3\) of H\(_2\)C\(_2\)O\(_4\) reacted completely. [0 = 16.0, K= 39.0, Mn = 55.0, Molar volume of gas at s.t.p.= 22.4 dm\(^3\) mol\(^{-1}\)]
Credit will be given for strict adherence to the instructions. for observations precisely recorded and jor accurate inferences. All tests, observations and inferences must be clearly entered in your answer book in ink, at the time they are made
| Burette Reading (cm\(^3\)) | Rough | 1st Titration | 2nd Titration | 3rd Titration |
| Final Burette Readings (cm\(^3\)) | 36.50 | 30.40 | 41.70 | 26.80 |
| Initial Burette readings (cm\(^3\)) | 12.30 | 05.90 | 17.10 | 01.90 |
| Vol. of acid used (cm\(^3\)) | 24.20 | 24.50 | 24.60 | 24.90 |
Titre value = \(\frac{24.50 + 24.60}{2}\) = 24.55cm\(^3\)
(b)(i) Molar concentration of KMNO\(_4\)
\(\frac{C(MnO_4^-) \times V(MnO_4^-)}{C(C_2O_4^{2-}) \times V(C_2O_4^{2-})} = \frac{n(MnO_4^-)}{n(C_2O_4^{2-})}\)
\(\frac{C(MnO_4^-) \times 24.55}{0.05 \times 25} = \frac{2}{5}\)
C(MnO\(_4^-\)) = \(\frac{0.05 \times 25 \times 2}{24.55 \times 5}\)
= 0.0204 mol/dm
(ii) Mass concentration of KmnO\(_4\)
Molar mass of KmnO\(_4\) = 39 + 55 + (16 x 4) = 158g/mol
Mass conc. KmnO\(_4\) = molar conc. x molar mass
= 0.0204 x 158 = 3.22g/dm\(^3\)
(iii) Volume of CO\(_2\) (gas evolved)
\(\frac{n(CO_2)}{n (H_2C_2O_4)}\) = \(\frac{10}{5}\)
n(CO\(_2\)) = \(\frac{10}{5}\) x 0.05
= 2 x 0.05 = 0.100 mol
V(CO\(_2\)) at s.t.p = n x Vm
25cm\(^3\) of H\(_2\)C\(_2\)O\(_4\) = 0.100 x 22.4dm\(^3\)mol\(^{-1}\) x \(\frac{25}{1000}\)
= 0.056dm\(^3\) or 56.0 cm\(^3\)
Alternative method
Volume of CO\(_2\) gas evolved
Amount of H\(_2\)C\(_2\)O\(_4\) used = 0.050 x \(\frac{25}{1000}\) = 0.00125 mol
From the equation of the reaction
5 mol of C\(_2\)O\(_4^{2-}\) = 10 mol of CO\(_2\)
0.00125 mol C\(_2\)O\(_4^{2-}\) = 0.00125 mol x \(\frac{10}{5}\)
= 0.0025 mol CO\(_2\)
@S.T.P
1 mol of CO\(_2\) = 22.4dm\(^3\)
0.00250 mol = 22.4 x 0.0025
= 0.056 dm\(_3\) or 56.0 cm\(^3\)
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