TEST OF PRACTICAL KNOWLEDGE QUESTION
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\(^{3}\) \(\mathrm{H_2C_2O_4}\) (ethanedioic acid). B is a solution of \(\mathrm{KMnO_4}\), (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 \(\mathrm{H_2SO_4}\), 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;
\[
\mathrm{2MnO}_{4(aq)}^{-} + \mathrm{5C_2O}_{4(aq)}^{2-} + \mathrm{16H}^{+}_{(aq)} \to \mathrm{2MnH}^{2+}_{(aq)} + \mathrm{8H_2O}_{(l)} + \mathrm{10CO}_{2(g)}
\]
(b) From your results and the information provided, calculate the:
(i) concentration of \(\mathrm{MnO_2^-}\) in B in moldm\(^{-1}\)
(ii) concentration of \(\mathrm{KMnO_4^-}\) in B in gdm\(^{-3}\)
(iii) volume of \(\mathrm{CO_2}\) evolved at s.t.p when 25.0 cm\(^{3}\) of \(\mathrm{H_2C_2O_4}\) 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
This is a redox titration of standard ethanedioic acid A (\(0.050\ \text{mol dm}^{-3}\)) against potassium tetraoxomanganate(VII) B of unknown concentration, using a 25.0 cm\(^3\) pipette of A.
\[2MnO_4^-+5C_2O_4^{2-}+16H^+\rightarrow 2Mn^{2+}+8H_2O+10CO_2\]
(a) Table of results.
| Burette reading (cm\(^3\)) | Rough | 1st | 2nd | 3rd |
|---|
| Final reading | 36.50 | 30.40 | 41.70 | 26.80 |
| Initial reading | 12.30 | 05.90 | 17.10 | 01.90 |
| Volume of B used | 24.20 | 24.50 | 24.60 | 24.90 |
Average of the two concordant titres:
\[V_B=\frac{24.50+24.60}{2}=24.55\ \text{cm}^3.\]
(b)(i) Concentration of \(MnO_4^-\) in B. Using the mole ratio \(n(MnO_4^-):n(C_2O_4^{2-})=2:5\):
\[\frac{C(MnO_4^-)\times24.55}{0.050\times25.0}=\frac{2}{5}\]\[C(MnO_4^-)=\frac{0.050\times25.0\times2}{24.55\times5}=0.0204\ \text{mol dm}^{-3}.\]
(ii) Concentration of \(KMnO_4\) in g dm\(^{-3}\). Molar mass of \(KMnO_4=39.0+55.0+(4\times16.0)=158\ \text{g mol}^{-1}\):
\[C=0.0204\times158=3.22\ \text{g dm}^{-3}.\]
(iii) Volume of \(CO_2\) at s.t.p. from 25.0 cm\(^3\) of A.
\[n(C_2O_4^{2-})=0.050\times\frac{25.0}{1000}=0.00125\ \text{mol}.\]
From the equation, 5 mol \(C_2O_4^{2-}\) give 10 mol \(CO_2\):
\[n(CO_2)=0.00125\times\frac{10}{5}=0.00250\ \text{mol}.\]\[V(CO_2)=0.00250\times22.4=0.056\ \text{dm}^3=56.0\ \text{cm}^3.\]
This is a redox titration of standard ethanedioic acid A (\(0.050\ \text{mol dm}^{-3}\)) against potassium tetraoxomanganate(VII) B of unknown concentration, using a 25.0 cm\(^3\) pipette of A.
\[2MnO_4^-+5C_2O_4^{2-}+16H^+\rightarrow 2Mn^{2+}+8H_2O+10CO_2\]
(a) Table of results.
| Burette reading (cm\(^3\)) | Rough | 1st | 2nd | 3rd |
|---|
| Final reading | 36.50 | 30.40 | 41.70 | 26.80 |
| Initial reading | 12.30 | 05.90 | 17.10 | 01.90 |
| Volume of B used | 24.20 | 24.50 | 24.60 | 24.90 |
Average of the two concordant titres:
\[V_B=\frac{24.50+24.60}{2}=24.55\ \text{cm}^3.\]
(b)(i) Concentration of \(MnO_4^-\) in B. Using the mole ratio \(n(MnO_4^-):n(C_2O_4^{2-})=2:5\):
\[\frac{C(MnO_4^-)\times24.55}{0.050\times25.0}=\frac{2}{5}\]\[C(MnO_4^-)=\frac{0.050\times25.0\times2}{24.55\times5}=0.0204\ \text{mol dm}^{-3}.\]
(ii) Concentration of \(KMnO_4\) in g dm\(^{-3}\). Molar mass of \(KMnO_4=39.0+55.0+(4\times16.0)=158\ \text{g mol}^{-1}\):
\[C=0.0204\times158=3.22\ \text{g dm}^{-3}.\]
(iii) Volume of \(CO_2\) at s.t.p. from 25.0 cm\(^3\) of A.
\[n(C_2O_4^{2-})=0.050\times\frac{25.0}{1000}=0.00125\ \text{mol}.\]
From the equation, 5 mol \(C_2O_4^{2-}\) give 10 mol \(CO_2\):
\[n(CO_2)=0.00125\times\frac{10}{5}=0.00250\ \text{mol}.\]\[V(CO_2)=0.00250\times22.4=0.056\ \text{dm}^3=56.0\ \text{cm}^3.\]
