All your burette readings (initial and final), as well as the size of your pipette, must be recorded but on no account of experiment procedure is required. ...
Question 1 Report
All your burette readings (initial and final), as well as the size of your pipette, must be recorded but on no account of experiment procedure is required. All calculations must be done in your answer book.
A is mol dm HCI. B is a solution containing 15.0 g dm of a mixture of NaCl and KHCO\(_3\).
(a) Put A burette and titrate it against \(20.0\text{cm}^3\) or \(25.0\text{cm}^3\) portions of B using methyl orange as indicator. Record the volume of your pipette. Tabulate your burette readings and calculate the average volume of A used. The equation for the reaction involved in the titration is:
\[ \mathrm{HCl}_{aq} + \mathrm{KHCO}_{3(aq)} \to \mathrm{KCl}_{(aq)} + \mathrm{CO}_{2(g)} \](b) From your results and the information provided above, calculate the:
(i) concentration of KHCO\(_3\), in \(\text{mol dm}^{-3}\) in B;
(ii) mass of KHCO\(_3\), in \(\text{g dm}^{-3}\) in B
(ii) Percentage by mass of KHCO\(_3\) in the mixture, [H=1; C = 12; O = 16; K = 39]
(iv) mass of NaCl in the mixture.
Answer Details
Titration results
Use a 25.0 cm3 pipette to measure solution B. The rough titre is not included in the average. The concordant titres are 26.00 cm3, 26.05 cm3 and 25.95 cm3.
| Reading | Rough | 1st accurate | 2nd accurate | 3rd accurate |
|---|---|---|---|---|
| Final burette reading / cm3 | 27.00 | 36.00 | 26.05 | 40.95 |
| Initial burette reading / cm3 | 0.00 | 10.00 | 0.00 | 15.00 |
| Volume of A used / cm3 | 27.00 | 26.00 | 26.05 | 25.95 |
Volume of pipette \(=25.0\ \text{cm}^3\).
Average volume of A used:
\[ \frac{26.00+26.05+25.95}{3}=26.00\ \text{cm}^3 \]
The equation must include water, since hydrogen carbonate reacts with acid to form carbon dioxide and water:
\[ \mathrm{HCl_{(aq)}+KHCO_{3(aq)}\rightarrow KCl_{(aq)}+CO_{2(g)}+H_2O_{(l)}} \]
Only \(\mathrm{KHCO_3}\) reacts with hydrochloric acid. Sodium chloride does not react, so it does not affect the titre. The mole ratio of \(\mathrm{HCl}\) to \(\mathrm{KHCO_3}\) is \(1:1\).
(i) Concentration of \(\mathrm{KHCO_3}\) in B
Using the stated concentration of A, \(0.100\ \text{mol dm}^{-3}\):
\[ n(\mathrm{HCl})=CV \]
\[ =0.100\times\frac{26.00}{1000} =2.60\times10^{-3}\ \text{mol} \]
Therefore, by the \(1:1\) ratio:
\[ n(\mathrm{KHCO_3})=2.60\times10^{-3}\ \text{mol} \]
This amount is present in \(25.0\ \text{cm}^3=0.0250\ \text{dm}^3\) of B:
\[ [\mathrm{KHCO_3}] =\frac{2.60\times10^{-3}}{0.0250} =\boxed{0.104\ \text{mol dm}^{-3}} \]
(ii) Mass concentration of \(\mathrm{KHCO_3}\) in B
\[ M_r(\mathrm{KHCO_3})=39+1+12+(3\times16)=100 \]
\[ \text{Mass concentration} =0.104\times100 =\boxed{10.4\ \text{g dm}^{-3}} \]
(iii) Percentage by mass of \(\mathrm{KHCO_3}\) in the mixture
The total mass concentration of the mixture is \(15.0\ \text{g dm}^{-3}\), of which \(10.4\ \text{g dm}^{-3}\) is \(\mathrm{KHCO_3}\):
\[ \%\,\mathrm{KHCO_3} =\frac{10.4}{15.0}\times100 =\boxed{69.3\%} \]
(iv) Mass of \(\mathrm{NaCl}\) in the mixture
\[ \text{Mass concentration of NaCl} =15.0-10.4 =\boxed{4.6\ \text{g dm}^{-3}} \]
The percentage calculation must use \(\frac{\text{mass of KHCO}_3}{\text{total mass of mixture}}\times100\). Using \(15.0-10.4\) in this fraction would calculate the percentage of sodium chloride instead, not the percentage of potassium hydrogencarbonate.
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