Burette reading (initial and final) must be given to two decimal places. Volume of pipette used must be recorded but no account of experimental procedure is required. All calculations must be done in you answer book.
A is a solution containing 6.3 g dm\(^{-3}\) of HNO\(_3\), B is a solution Na\(_2\)CO\(_3\)
(a) Put A Into the burette and titrate it against 20.0 cm\(^3\) portions of B using methyl orange indicator. Record the volume of your pipette. Repeat the titration to obtain consistent titres. Tabulate your burette readings and calculate the average volume of A used. The equation for the reaction involved in the titration is:
2HNO\(_{3(aq)}\) + Na\(_2\)CO\(_{3(aq)}\) \(\to\) 2NaNO\(_{3(aq)}\) + CO\(_{2(g)}\) + H\(_{2}\)O\(_{(l)}\)
(b) From your results and information provided above, calculate the;
(i) concentration of B in mol dm\(^{-3}\)
(ii) concentration of B in g dm\(^{-3}\):
(iii) mass of sodium ions in 1:0 dm\(^{-3}\) of B
[H = 1; C = 1; O = 16; N = 14; Na = 23]
Explanation
(a)
| Titrations/Burette reading | Rough Titration (cm\(^3\)) | 1st Titration (cm\(^3\)) | 2nd Titration (cm\(^3\)) | 3rd Titration (cm\(^3\) |
| Final | 22.70 | 45.30 | 22.60 | 45.20 |
| Initial | 0.00 | 22.70 | 22.70 | 0.00 |
| Volume of acid used | 22.70 | 22.60 | 22.60 | 22.60 |
Volume of pipette used 25.0 cm\(^{3}\)
2HNO\(_{3(aq)}\) + Na\(_2\)CO\(_{3(aq)}\) \(\to\) 2NaNO\(_{3(aq)}\) + CO\(_{2(g)}\) + H\(_2\)O\(_{(l)}\)
Average volume of acid used
= 22.60 + \(\frac{22.60}{3}\) + 22.60
= \(\frac{67.80}{3}\)
= 22.60 cm\(^3\)
(b)(i) Molar mass of HNO\(_3\) = 1 + 14 + 48
= 63 g/mol
concentration of A = \(\frac{6.3}{63}\)
= 0.1 mol/dm\(^3\)
To calculate the concentration of solution B;
\(\frac{C_AV_A}{C_BV_B} = \frac{N_A}{N_B}\)
C\(_A\) = 0.1 mol/dm\(^3\)
C\(_B\) = ?
N\(_A\) = 2
V\(_A\) = 22.60 cm\(^3\)
V\(_B\) = 25.0 cm\(^3\)
N\(_B\) = 1
\(\frac{0.1 \times 22.60}{C_B \times 25} = \frac{2}{1}\)
C\(_B\) = \(\frac{0.1 \times 22.60}{2 \times 25} = \frac{2.26}{50}\) = 0.0452
= 0.0450 mol/dm\(^3\)
(ii) Na\(_2\)CO\(_3\) = 23 x 2 + 12 + 16 x 3 = 46 + 48
= 106 g/mol
Concentration of B in g/dm\(^3\) = 4.77 g/dm\(^3\)
(iii) To calculate mass of Na\(^{+}\) in 1.0 dm\(^3\)
Concentration of N\(^+\) in mol/dm\(^3\) = 2 x 0.0450
= 0.09 mol/dm\(^3\)
Since Na,
Mass of Na\(^+\) in 1.0 dm\(^3\) = 23 x 0.09 = 2.07
(iii) C is a mixture of starch and (NH\(_4\))\(_2\)SO\(_4\)