All your burette readings (initial and final) as well as the size size of your pipette, must be recorded but no account of experimental procedure is required. All calculations must be done in your answer book.
Xg of pure potassium trioxocarbonate (IV) was treated with 1dm\(^3\) of 0. 25M tetraoxosulphate (VI) acid to obtain solution A which contains excess acid. B is a solution containing 2.8g of potassium hydroxIde per 250cm\(^3\) solution.
(a) Put A into the burette and titrate with 20cm\(^3\) or 25cm\(^3\) portions of B. using methyl orange as indicator. Record the volume of your pipette. Tabulate your burette readings and calculate the average volume of acid used.
(b) From your results and the information given. Calculate the;
(i) concentration of the excess acid in A in mol dm\(^3\)
(iii) value of X. The equation for the reaction between the excess acid the potassium hydroxide is H\(_2\)SO\(_4\) + 2KOH \(\to\) K\(_2\)SO\(_4\) [H = 1. C = 12, O = 16. S = 32, K = 39]
Volume of pipette = 25.00 cm\(^3\)
Indicator used - methyl orange
Colour change at the end profit - from yellow/pink
| Burette reading | Rough | 1st | 2nd | 3rd |
| Final reading in cm\(^3\) | 24.60 | 24.00 | 23.95 | 27.00 |
| Initial reading in cm\(^3\)0.00 | 0.0.00 | 0.00 | 0.00 | 3.00 |
| Difference in cm\(^3\) | 24.60 | 24.00 | 23.95 | 24.05 |
Average volume of acid used = \(\frac{24.00 + 23.95 + 24.05}{3}\)
= 24.00cm\(^3\)
(b)(i) B contain 2.8g of KOH per 250cm\(^3\) (given)
Molar mass concentration of B = \(\frac{2.8 \times 100}{250}\) = 11.2 g/dm\(^3\)
molar mass of KOH = 56
Molar concentration of KOH = \(\frac{11.6}{56}\) = 0.20 mol/dm\(^3\)
Equation for reaction; 2KOH + H\(_2\)SO\(_4\) = K\(_2\)SO\(_4\) + 2H\(_2\)O
from \(\frac{\text{molarity of A(M\(_A\)) x volume of A(V\(_A\))}}{\text{molarity of B(M\(_B\)) x volume of A(V\(_B\))}}\) = mole ratio of \(\frac{A}{B}\)
M\(_A\) = ?
M\(_s\) = 0.20m
V\(_A\) = 24.00 cm\(^3\)
V\(_B\) = 25.00 cm\(^3\)
ratio of \(\frac{A}{B}\) = \(\frac{1}{2}\)
\(\frac{M_A \times 24}{0.20 \times 25} = \frac{1}{2}\)
M\(_A\) = \(\frac{0.20 \times 25}{24 times 2}\)
= 0.1042 mol dm\(^{-3}\)
The molarity of excess acid = 0.1042m;
Amount of acid in mol dm\(^{-3}\) that reacted = 0.2500 - 0.1042 = 0.1458 mol.
(ii) H\(_2\)SO\(_4\) + K\(_2\)CO\(_3\) = K\(_2\)SO\(_4\) + H\(_2\)O + CO\(_2\) = 148
value of x = 138 x 0.1458 = 20.12g
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