Burette readings (initial and final) must be given to two decimal places. Volume of pipefte used must also be recored but no account of expeririental procedure is required. All calculations must be done in your answer book.
A is 0.100 mol dm\(^{-3}\) solution of an acid. B is a solution of KOH containing 2.8 g per 500 cm’\(^3\)
(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 an indicator. Repeat the titration to obtain consistent titres. Tabulate your readings and calculate the average volume of A used.
(b) From your results and the information provided above, calculate the:
(i) number of moles of acid in the average titre;
(i) number of moles of KOH in the volume of B pipetted;
(ii) mole ratio of acid to base in the reaction. [H = 1.00, O = 16.0, K = 39.0]
Explanation
| Rough cm(\(^3\)) | 1st Titration (cm\(^3\)) | 2nd Titration (cm\(^3\)) | |
| Final Burette Readings | 24.10 | 47.00 | 23.10 |
| Initial Burette Readings | 01.10 | 24.10 | 00.00 |
| Volume of Acid used | 23.00 | 22.90 | 23.10 |
Average Titre = \(\frac{23.00 + 22.90 + 23.100}{3}\)
= 23.0 cm\(^3\)
(b)(i) No of moles of acid = \(\frac{0.100 \times 23.00}{1000}\)
= 0.00230 moles
(ii) No of moles of KOH in B
500cm\(^3\) of B contains 2.8g of KOH
1000cm\(^3\) of B will contain \(\frac{2.8 \times 1000}{500}\)
= 5.60g KOH
Molar mass of KOH = 39 + 16 + 1
= 56 g mcl\(^{-1}\)
Conc. of B = \(\frac{5.6}{56} = 0.100 mol dm^{-3}\)
Number of moles of KOH in
B = \(\frac{0.100 \times 25}{1000}\)
= 0.00250 moles
(ii) Mole ratio of acid to base = 0.0023 : 0.0025
= 1 : 1