TEST OF PRACTICAL KNOWLEDGE QUESTION
Burette readings (initial and final reading) must be given to two decimal places. Volume of pipette used must also be recorded but no account of experimental procedure is required. All calculations must be done in your answer book.
A is 0.0950 mol dm\(^{-3}\) HCI. B is a solution 13.50g dm\(^{-3}\) of X\(_2\)CO\(_3\).10H\(_2\)O.
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^3\) portions öf B using methyl orange as an indicator. Tabulate your readings and calculate the average volume of A used.
(b) From your results and the information provided above, calculate the;
(i) concentration of B in mol dm\(^{-3}\);
(ii) molar mass of X\(_2\)CO\(_3\).10H\(_2\)O in g mol\(^{-1}\);
(iii) percentage by mass X in X\(_2\)C)\(_3\).10H\(_2\)O. [H = 1, C = 12, O = 16]. The equation for the reaction involved in the titration is 2HCl\(_{(aq)}\) + X\(_2\)CO\(_3\).10H\(_2\)O\(_{(aq)}\) \(\to\) 2XCl\(_{(aq)}\) + 11H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)
(a) Titration
A (0.0950 mol dm-3 HCl) is placed in the burette and 25.0 cm3 portions of B are pipetted into the conical flask with methyl orange as indicator. A specimen set of concordant readings is shown.
| Burette readings (cm3) | Rough | 1st | 2nd |
|---|
| Final reading | 25.10 | 24.90 | 24.85 |
| Initial reading | 0.00 | 0.00 | 0.00 |
| Volume of A used | 25.10 | 24.90 | 24.85 |
Average volume of A (concordant titres only) \(=\dfrac{24.90+24.85}{2}=24.88\ \text{cm}^3\)
(b)(i) Concentration of B in mol dm-3
\[ n(\text{HCl})=\frac{0.0950\times 24.88}{1000}=2.364\times10^{-3}\ \text{mol} \]
From the equation, HCl : X2CO3.10H2O = 2 : 1, so
\[ n(B)=\frac{2.364\times10^{-3}}{2}=1.182\times10^{-3}\ \text{mol in }25.0\ \text{cm}^3 \]
\[ [B]=\frac{1.182\times10^{-3}\times1000}{25.0}=0.0473\ \text{mol dm}^{-3} \]
(ii) Molar mass of X2CO3.10H2O
Since concentration in g dm-3 = molar mass \(\times\) concentration in mol dm-3:
\[ M=\frac{13.50}{0.0473}=285.4\approx 286\ \text{g mol}^{-1} \]
(iii) Percentage by mass of X in X2CO3.10H2O
Mass of the (CO3 + 10H2O) part \(= 12 + (3\times16) + 10(18) = 60 + 180 = 240\).
So mass of 2X \(= 286 - 240 = 46\), giving X = 23 (sodium).
\[ \%X=\frac{46}{286}\times100=16.1\% \]
The titre figures above are specimen readings; use your own concordant burette values. The result is consistent with X = Na, i.e. washing soda, Na2CO3.10H2O.
(a) Titration
A (0.0950 mol dm-3 HCl) is placed in the burette and 25.0 cm3 portions of B are pipetted into the conical flask with methyl orange as indicator. A specimen set of concordant readings is shown.
| Burette readings (cm3) | Rough | 1st | 2nd |
|---|
| Final reading | 25.10 | 24.90 | 24.85 |
| Initial reading | 0.00 | 0.00 | 0.00 |
| Volume of A used | 25.10 | 24.90 | 24.85 |
Average volume of A (concordant titres only) \(=\dfrac{24.90+24.85}{2}=24.88\ \text{cm}^3\)
(b)(i) Concentration of B in mol dm-3
\[ n(\text{HCl})=\frac{0.0950\times 24.88}{1000}=2.364\times10^{-3}\ \text{mol} \]
From the equation, HCl : X2CO3.10H2O = 2 : 1, so
\[ n(B)=\frac{2.364\times10^{-3}}{2}=1.182\times10^{-3}\ \text{mol in }25.0\ \text{cm}^3 \]
\[ [B]=\frac{1.182\times10^{-3}\times1000}{25.0}=0.0473\ \text{mol dm}^{-3} \]
(ii) Molar mass of X2CO3.10H2O
Since concentration in g dm-3 = molar mass \(\times\) concentration in mol dm-3:
\[ M=\frac{13.50}{0.0473}=285.4\approx 286\ \text{g mol}^{-1} \]
(iii) Percentage by mass of X in X2CO3.10H2O
Mass of the (CO3 + 10H2O) part \(= 12 + (3\times16) + 10(18) = 60 + 180 = 240\).
So mass of 2X \(= 286 - 240 = 46\), giving X = 23 (sodium).
\[ \%X=\frac{46}{286}\times100=16.1\% \]
The titre figures above are specimen readings; use your own concordant burette values. The result is consistent with X = Na, i.e. washing soda, Na2CO3.10H2O.
