Burette readings(initial and final) must be given to two decimal places. Volume of pipette user must also be recorded but on account of experimental procedure is required. All calculations must be done in your answer book. A is O.050 mol dm\(^{-3}\) of acid HX. Bis a solution of NaOH containing 0.025 moles per 250 solutions.
(a) Put A into the burette and titrate it against 20.00 cm\(^3\) or 25.00 cm\(^3\) portions B using phenolphthalein as indicator. Tabulate your readings and calculate the average volume or A used.
(b) your results and the information provided above, calculate the;
(i) amount of acid in the average
(ii) amount of base in 20.00 cm\(^3\) or 25.00 cm\(^3\);
(iii) mole ratio of acid to base
(c) Write a balanced chemical equation for the reaction between the acid H\(_y\)X and the base NaOH
(d) State the basicity of the acid H\(_y\)X.
Explanation
Table of Result
Indicator used: Phenolphthalein
Volume of pipette = 25 cm\(^3\)
| Burette Reading | Rough | 1st | 2nd |
3rd
|
| Final (cm\(^3\)) |
22.60
|
22.40 | 22.30 |
22.30 |
| Initial (cm\(^3\)) |
0.00
|
0.00 | 0.00 | 0.00 |
| Volume of acid used (cm\(^3\)) |
22.00
|
22.40 | 22.30 | 22.30 |
Average volume of acid used, V\(_A\) = \(\frac{22.40 + 22.30 + 22.30}{3}\)
= \(\frac{67.00}{3}\)
Average titre value = 22.30 cm\(^3\), V\(_A\)
(b)(i) Amount of acid in the average titre, V\(_A\)
1000 cm\(^3\) contains 0.05 moles
V\(_A\) will contain = \(\frac{0.05 \times 22.30}{1000}\)
= 0.0011 moles
(ii) Concentration of NaOH in B
\(\frac{0.025}{0.25}\) = 0.199 mol dm\(^{-3}\)
To calculate the amount of base in 25 cm\(^3\)
\(\frac{25 cm^3}{1000}\) x 0.100 mol dm\(^{-3}\) = 0.0025 moles
(iii) Mole ratio of acid (H\(_y\)X) to base (NaOH)
\(\frac{0.0011}{0.0025}\) = \(\frac{1}{2}\)
(c) Equation for the reaction
2NaOH + H\(_y\)X \(\to\) Na\(_2\)X + H\(_2\)O
(d) Basicity of acid H\(_y\)X = 2, (two)