All your burette readings (initial and final), as well as the size of your pipette, must be recorded but on no account of experimental procedure is required. All calculations must be done in your answer book.
A is solution of trioxonitrate (V) acid, B is a solution containing 6.90 g of potassium trioxocarbonate (IV) per dm\(^3\)
(a) Put A into the buret and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^3\) portions of B using methy orange or screened methyl orange as indicater. Record the volume of your pipette. Tabulate your burette readings and calculate the average volume A used. The equation for the reaction is K\(_2\)CO\(_{3(aq)}\) + 2HNO\(_{3(aq)}\) \(\to\) 2KNO\(_{3(aq)}\) + CO\(_{2(g)}\) + H\(_2\)O\(_{(l)}\)
(b) From your results and the information provided calculate;
(i) concentration of solution B in mol dm\(^{-3}\)
(ii) number of potassium ions in 1.00 dm\(^3\) of B [C = 12.0, O = 16.0, K = 39.0, Avogadro constant = 6.02 x 10\(^{23}\) mol \(^{-1}\)]
Explanation
(a) Volume of pipette 25.00 cm\(^3\)
Indicator used - Methyl orange, colour changed at end point - from yellow to orange/pink
| Burette readings | Rough | 1st | 2nd | 3rd |
| Final reading in cm\(^3\) | 16.50 | 16.00 | 15.95 | 26.05 |
| Initial reading in cm\(^3\) | 0.00 | 0.00 | 0.00 | 10.00 |
| Difference in cm\(^3\) | 16.50 | 16.00 | 15.95 | 16.05 |
Average volume of acid used = \(\frac{16.00 + 15.95 + 16.05}{3}\) = 16.00 cm\(^3\)
Equation for the reaction K\(_2\)CO\(_3\) + 2HNO\(_3\) → 2KNO\(_3\) + CO\(_2\) + H\(_2\)O
Molar mass of K\(_2\)CO\(_3\) = 2 x 39 + 2 + 16 x 3 = 138g
No. of moles of B = \(\frac{6.90}{138}\) = 0.0500 mol/dm\(^3\)
(ii) 1 mole of B contains 6.02 x 10\(^{23}\) mol\(^{-1}\)
0.0500 moles = 6.02 x 10\(^{23}\) x 0.05
= 3.01 x 10\(^{22}\)
The number of potassium ions per dm\(^3\) = 3.01 x 10\(^{22}\)
(iii) \(\frac{\text{mole concentration of A x volume of A}}{\text{mole concentration of B x volume of B}}\) = mole ratio of \(\frac{A}{B}\)
\(\frac{MA \times 16.00}{0.05 \times 25.00} = \frac{2}{1}\)
MA = \(\frac{0.05 \times 25.00 \times 2}{16.00}\) = 0.1563 mol/dm\(^3\) (4d.p)