Burette readings (initials and final) must be given to two decimal places. Volume of pipette used must also be recorded but no account experimental procedure is required. Al calculations must be done in your answer book. A solution containing 0.05moldm\(^3\) H\(_2\)SO\(_4\). B is a solution containing 1.4g per 250cm\(^3\) .XOH
(a) Put A into the burette and titrate it against 20.0cm\(^3\) or 25.0cm\(^3\) portions of B using methyl orange indicator. Repeat the titration to obtain consistent titres. Tabulate your results and calculate the average volume of A used. The equation for the reaction involved in the titration is; H\(_2\)SO\(_{4(aq)}\) + 2XOH\(_{(aq)}\) \(\to\) X\(_2\)SO\(_{4(aq)}\) + 2H\(_2\)O\(_{(l)}\)
(b) From your results and the information provided above, calculate the;
(i) concentration of B in moldm\(^{-3}\)
(ii) molar mass XOH
(iii) relative atomic mass of X. [H = 1.00; O = 16.0 S =32.0]
Explanation
(a)
| Burette reading | Rough titration (cm\(^3\)) | 1st titration (cm\(^3\)) | 2nd titration (cm\(^3\)) |
| Final Burette reading | 21.50 | 37.10 | 22.60 |
| Initial Burette reading | 06.00 | 21.50 | 07.10 |
| 15.50 | 15.60 | 15.50 |
Average titre = \(\frac{15.50 + 15.60 + 15.50}{3}\)
= 15.53 cm\(^3\)
(b)(i) Concentration of B in moldm\(^{-3}\)
H\(_2\)SO\(_4\) + 2XOH \(\to\) X\(_2\)SO\(_4\) + 2H\(_2\)O
\(\frac{C_AV_A}{C_BV_B}\) = \(\frac{1}{2}\)
C\(_B\) = \(\frac{C_AV_A \times 2}{V_B}\)
= \(\frac{0.05 \times 15.53 \times 2}{25}\)
= 0.0621 moldm\(^{-3}\)
(ii) Molar mass of XOH
Conc. B in g/dm\(^3\) = \(\frac{1.4 \times 1000}{250}\)
= 5.6 g/dm\(^3\)
Molar mass of XOH = \(\frac{5.6 g/dm^3}{0.0621 mol/dm^3}\)
= 9.2 g/mol
Sample C is Fe(NH\(_4\))\(_2\) (SO\(_4\))\(_2\)
(ii) Relative atomic mass of X
XOH = 90.2
X + 16 + 1 = 90.2
X = 90.2 - 17
= 73.2