All your burette readings (initial and final), as well as the size of your pipette, must be recorded but no account or expeimental procedure is required. All calculations must be done in your answer book.
A is 0.125 mol dm\(^3\) H\(_2\)SO\(_4\). B is a solution containing X g dm\(^{-3}\) of NaOH.
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 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 H\(_2\)SO\(_4\) +2NaOH\(_{(aq)}\) \(\to\) Na\(_2\)SO\(_4\) + 2H\(_2\)O\(_{(l)}\)
(b) From your results and the information provided above, calculate the;
(i) amount of H\(_2\)SO\(_4\) in the average volume of A used
(ii) Concentration of B in mol dm\(^{-3}\)
(iii) value of X.
[H = 1: O = 16; Na = 23]
(c) Describe briefly a suitable laboratory procedure for obtaining pure water from the titration mixture. (No diagram is required)
Explanation
(a) Volume of pipette = 25.00 cm\(^3\)l indicator used = methyl orange, colour change at endpoint from yellow to orange/purple
| Burette reading | Rough | 1st | 2nd |
3rd
|
|
Final burette readings(cm\(^3\)) |
24.00 | 23.25 | 24.15 | 38.20 |
| Final burette readings (cm\(^3\)) | 0.00 | 0.00 | 10.00 | 15.00 |
| Volume of A used. | 24.00 | 23.25 | 23.15 | 23.20 |
Average volume of A used = \(\frac{23.25 + 23.15 + 23.20}{3}\) = \(\frac{69.6}{3}\) = 23.20 cm\(^3\)
(b)(i) A contains 0.125 mol/dm\(^3\) of H\(_2\)SO\(_4\) (given)
I.e. 0.125 in 1000 cm\(^3\) of H\(
_2\)SO\(_4\)
x in 23.20; x = \(\frac{23.20}{1000}\) x 0.125 = 0.0029 mol per 23.20 cm\(^3\)
(ii) \(\frac{\text{concentration of A x volume of A}}{\text{concentration of B x volume of B}}\) = mole ratio of \(\frac{A}{B}\)
C\(_B\) = \(\frac{0.125 \times 23.20}{C_B \times 25}\) = \(\frac{1}{2}\)
C\(_B\) = 0.232 mol/dm\(^3\)
(iii) B contains Xg/dm\(^3\) of NaOH, molar mass of NaOH = 23 + 16 + 1 = 40
mass concentration = mole concentration x molar mass = 0.2332 x 40
x = 0.28g
(c) Pure water can be obtained from the solution of Na\(_2\)SO\(_4\) + 2H\(_2\)O by ordinary distillation. Distillation involves the boiling point only and the vapour condensed to give the distllate.