TEST OF PRACTICAL KNOWLEDGE QUESTION
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.
(a) Table of results. Indicator: phenolphthalein. Volume of pipette = 25.00 cm\(^3\). A (acid HX) is in the burette, titrated against 25.00 cm\(^3\) portions of B (NaOH).
| Burette reading (cm\(^3\)) | Rough | 1st | 2nd | 3rd |
|---|
| Final reading | 22.60 | 22.40 | 22.30 | 22.30 |
| Initial reading | 0.00 | 0.00 | 0.00 | 0.00 |
| Volume of A used | 22.60 | 22.40 | 22.30 | 22.30 |
Average of the concordant titres:
\[V_A=\frac{22.40+22.30+22.30}{3}=\frac{67.00}{3}=22.30\ \text{cm}^3.\]
(b)(i) Amount of acid in the average titre. A is \(0.050\ \text{mol dm}^{-3}\):
\[n_{acid}=\frac{0.050\times22.30}{1000}=0.0011\ \text{mol}.\]
(ii) Amount of base in 25.00 cm\(^3\) of B. Concentration of NaOH:
\[[\text{NaOH}]=\frac{0.025}{250/1000}=\frac{0.025}{0.250}=0.10\ \text{mol dm}^{-3}.\]\[n_{base}=\frac{0.10\times25.00}{1000}=0.0025\ \text{mol}.\]
(iii) Mole ratio of acid to base.
\[\frac{n_{acid}}{n_{base}}=\frac{0.0011}{0.0025}=\frac{1}{2},\qquad\text{acid : base}=1:2.\]
(c) Balanced equation. The acid combines with two moles of NaOH, so it is H\(_2\)X:
\[H_2X+2NaOH\rightarrow Na_2X+2H_2O.\]
(d) Basicity of the acid. The acid reacts with NaOH in the ratio 1 : 2, so it has two replaceable hydrogen atoms. Basicity = 2.
(a) Table of results. Indicator: phenolphthalein. Volume of pipette = 25.00 cm\(^3\). A (acid HX) is in the burette, titrated against 25.00 cm\(^3\) portions of B (NaOH).
| Burette reading (cm\(^3\)) | Rough | 1st | 2nd | 3rd |
|---|
| Final reading | 22.60 | 22.40 | 22.30 | 22.30 |
| Initial reading | 0.00 | 0.00 | 0.00 | 0.00 |
| Volume of A used | 22.60 | 22.40 | 22.30 | 22.30 |
Average of the concordant titres:
\[V_A=\frac{22.40+22.30+22.30}{3}=\frac{67.00}{3}=22.30\ \text{cm}^3.\]
(b)(i) Amount of acid in the average titre. A is \(0.050\ \text{mol dm}^{-3}\):
\[n_{acid}=\frac{0.050\times22.30}{1000}=0.0011\ \text{mol}.\]
(ii) Amount of base in 25.00 cm\(^3\) of B. Concentration of NaOH:
\[[\text{NaOH}]=\frac{0.025}{250/1000}=\frac{0.025}{0.250}=0.10\ \text{mol dm}^{-3}.\]\[n_{base}=\frac{0.10\times25.00}{1000}=0.0025\ \text{mol}.\]
(iii) Mole ratio of acid to base.
\[\frac{n_{acid}}{n_{base}}=\frac{0.0011}{0.0025}=\frac{1}{2},\qquad\text{acid : base}=1:2.\]
(c) Balanced equation. The acid combines with two moles of NaOH, so it is H\(_2\)X:
\[H_2X+2NaOH\rightarrow Na_2X+2H_2O.\]
(d) Basicity of the acid. The acid reacts with NaOH in the ratio 1 : 2, so it has two replaceable hydrogen atoms. Basicity = 2.
