TEST OF PRACTICAL KNOWLEDGE QUESTION
Burette readings (initial and final) must be given to two decimal places. Volume of pipette used must also be recorded but no account of experimental procedure is required. All calculations must be done in your answer book.
A solution containing 6.22 g of an acid \( \mathrm{H}_2\mathrm{Y} \) per dm\(^3\)
B contains 3.90 g of NaOH per dm\(^3\) of solution.
(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 indicator. Repeat the titration to obtain consistent titres. Tabulate your burette readings and calculate the average volume of acid A used. The equation for the reaction involved in the titration is:
\[
\mathrm{H}_2{}_{(aq)} + 2\mathrm{NaOH}_{(aq)} \to \mathrm{NaY}_{(aq)} + 2\mathrm{H}_2\mathrm{O}_{(l)}
\]
[H = 1.00; O = 16.0; Na = 23.0]
(b) From your results and the information provided above, calculate the:
(i) The concentration of B in moldm\(^{-3}\)
(ii) concentration of A in moldm\(^3\)
(ii) molar mass of H\(_2\)Y.
(c) State whether the pH of each of the following solutions is lower than 7, greater than 7 or equal to 7. The
(I) solution A before titration
(ii) solution B before titration
(a) Burette readings and average titre
| Burette reading (cm3) | Rough | 1st titration | 2nd titration |
|---|
| Final reading | 27.00 | 49.50 | 36.00 |
| Initial reading | 01.00 | 26.00 | 12.40 |
| Volume of A used | 26.00 | 23.50 | 23.60 |
Average titre \(= \dfrac{23.50 + 23.60}{2} = 23.55\ \text{cm}^3\) of A, for a 25.0 cm3 portion of B.
Equation: \(H_2Y_{(aq)} + 2NaOH_{(aq)} \rightarrow Na_2Y_{(aq)} + 2H_2O_{(l)}\).
(b)(i) Concentration of B in mol dm-3
Molar mass of NaOH \(= 23 + 16 + 1 = 40\ \text{g mol}^{-1}\).
\(C_B = \dfrac{3.90}{40} = 0.0975\ \text{mol dm}^{-3}\).
(ii) Concentration of A in mol dm-3
From \(\dfrac{C_A V_A}{C_B V_B} = \dfrac{n_A}{n_B} = \dfrac{1}{2}\), with \(V_A = 23.55\ \text{cm}^3\), \(V_B = 25.0\ \text{cm}^3\):
\(C_A = \dfrac{C_B V_B}{2 V_A} = \dfrac{0.0975 \times 25.0}{2 \times 23.55} = 0.0518\ \text{mol dm}^{-3}\).
(iii) Molar mass of H2Y
\(M = \dfrac{\text{mass concentration}}{\text{molar concentration}} = \dfrac{6.22}{0.0518} = 120\ \text{g mol}^{-1}\).
(c) pH before titration
- (i) Solution A (the acid H2Y): pH is less than 7.
- (ii) Solution B (NaOH): pH is greater than 7.
(a) Burette readings and average titre
| Burette reading (cm3) | Rough | 1st titration | 2nd titration |
|---|
| Final reading | 27.00 | 49.50 | 36.00 |
| Initial reading | 01.00 | 26.00 | 12.40 |
| Volume of A used | 26.00 | 23.50 | 23.60 |
Average titre \(= \dfrac{23.50 + 23.60}{2} = 23.55\ \text{cm}^3\) of A, for a 25.0 cm3 portion of B.
Equation: \(H_2Y_{(aq)} + 2NaOH_{(aq)} \rightarrow Na_2Y_{(aq)} + 2H_2O_{(l)}\).
(b)(i) Concentration of B in mol dm-3
Molar mass of NaOH \(= 23 + 16 + 1 = 40\ \text{g mol}^{-1}\).
\(C_B = \dfrac{3.90}{40} = 0.0975\ \text{mol dm}^{-3}\).
(ii) Concentration of A in mol dm-3
From \(\dfrac{C_A V_A}{C_B V_B} = \dfrac{n_A}{n_B} = \dfrac{1}{2}\), with \(V_A = 23.55\ \text{cm}^3\), \(V_B = 25.0\ \text{cm}^3\):
\(C_A = \dfrac{C_B V_B}{2 V_A} = \dfrac{0.0975 \times 25.0}{2 \times 23.55} = 0.0518\ \text{mol dm}^{-3}\).
(iii) Molar mass of H2Y
\(M = \dfrac{\text{mass concentration}}{\text{molar concentration}} = \dfrac{6.22}{0.0518} = 120\ \text{g mol}^{-1}\).
(c) pH before titration
- (i) Solution A (the acid H2Y): pH is less than 7.
- (ii) Solution B (NaOH): pH is greater than 7.
