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]
(a) Burette readings and average titre
| Burette reading (cm3) | Rough | 1st titration | 2nd titration |
|---|
| Final reading | 24.10 | 47.00 | 23.10 |
| Initial reading | 01.10 | 24.10 | 00.00 |
| Volume of A used | 23.00 | 22.90 | 23.10 |
Average titre \(= \dfrac{23.00 + 22.90 + 23.10}{3} = 23.0\ \text{cm}^3\) of A. (A 25.0 cm3 portion of B is pipetted.)
(b)(i) Number of moles of acid A in the average titre
\(= \dfrac{0.100 \times 23.00}{1000} = 2.30\times10^{-3}\ \text{mol} = 0.00230\ \text{mol}\).
(ii) Number of moles of KOH in the volume of B pipetted
Molar mass of KOH \(= 39 + 16 + 1 = 56\ \text{g mol}^{-1}\).
500 cm3 of B contains 2.8 g, so 1000 cm3 contains \(\dfrac{2.8 \times 1000}{500} = 5.60\ \text{g}\).
Concentration of B \(= \dfrac{5.60}{56} = 0.100\ \text{mol dm}^{-3}\).
Moles of KOH in 25.0 cm3 of B \(= \dfrac{0.100 \times 25.0}{1000} = 2.50\times10^{-3}\ \text{mol} = 0.00250\ \text{mol}\).
(iii) Mole ratio of acid to base
\(= 0.00230 : 0.00250 = 1 : 1\).
(a) Burette readings and average titre
| Burette reading (cm3) | Rough | 1st titration | 2nd titration |
|---|
| Final reading | 24.10 | 47.00 | 23.10 |
| Initial reading | 01.10 | 24.10 | 00.00 |
| Volume of A used | 23.00 | 22.90 | 23.10 |
Average titre \(= \dfrac{23.00 + 22.90 + 23.10}{3} = 23.0\ \text{cm}^3\) of A. (A 25.0 cm3 portion of B is pipetted.)
(b)(i) Number of moles of acid A in the average titre
\(= \dfrac{0.100 \times 23.00}{1000} = 2.30\times10^{-3}\ \text{mol} = 0.00230\ \text{mol}\).
(ii) Number of moles of KOH in the volume of B pipetted
Molar mass of KOH \(= 39 + 16 + 1 = 56\ \text{g mol}^{-1}\).
500 cm3 of B contains 2.8 g, so 1000 cm3 contains \(\dfrac{2.8 \times 1000}{500} = 5.60\ \text{g}\).
Concentration of B \(= \dfrac{5.60}{56} = 0.100\ \text{mol dm}^{-3}\).
Moles of KOH in 25.0 cm3 of B \(= \dfrac{0.100 \times 25.0}{1000} = 2.50\times10^{-3}\ \text{mol} = 0.00250\ \text{mol}\).
(iii) Mole ratio of acid to base
\(= 0.00230 : 0.00250 = 1 : 1\).
