TEST OF PRACTICAL KNOWLEDGE QUESTION
Burette readings (initial and final) must be given to two decimal places. Volume or pipete used must also be recorded but no account of experimental procedure is required. All calculations must be done in your answer book.
A is a solution containing \(12.0\text{ g dm}^{-3}\) \(\mathrm{NaHSO}_4\) \(\mathrm{NaHSO}_4\) P is a solution containing \(\mathrm{NaOH}\)
(a) Put A into the burette and titrate it against \(20.0\text{ cm}^3\) or \(25.0\text{ m}^3\) portions of B using methyl orange as an indicator. Repeat the titration to obtain consistent titres. Tabulate your readings and calculate the average volume A used. The equation for the reaction involved in the titration is ;
\[
\mathrm{NaHSO}_{4(aq)} + \mathrm{NaOH}_{(aq)} \to \mathrm{Na}_2\mathrm{SO}_{4(aq)} + \mathrm{H}_2\mathrm{O}_{(l)}
\]
[H 1.00, O = 16.0; Na = 23.0, S = 32.0]
(b) From your results and the information provided above calculate the:
(i) concentration of A in \(\text{mol dm}^{-3}\)
(ii) concentration of B in \(\text{mol dm}^{-3}\)
(iii) mass of \(\mathrm{Na}^+\) formed in solution during the titration.
Reaction:
\[ NaHSO_4 + NaOH \to Na_2SO_4 + H_2O \quad (1:1) \]
(a) Titration results (indicator: methyl orange; pipette volume of B = 25.00 cm3)
| Burette readings (cm3) | Rough | 1st | 2nd |
|---|
| Final reading | 21.70 | 43.30 | 31.90 |
| Initial reading | 0.00 | 21.70 | 10.30 |
| Volume of A used | 21.70 | 21.60 | 21.60 |
Average volume of A (from concordant titres):
\[ V_A = \frac{21.60 + 21.60}{2} = 21.60\ \text{cm}^3 \]
(b)(i) Concentration of A in mol dm-3
Molar mass of NaHSO4 \(= 23.0 + 1.00 + 32.0 + (4 \times 16.0) = 120.0\ \text{g mol}^{-1}\).
\[ [A] = \frac{12.0}{120.0} = 0.100\ \text{mol dm}^{-3} \]
(b)(ii) Concentration of B in mol dm-3
Since the reaction is 1:1, \(\dfrac{C_A V_A}{C_B V_B} = 1\), with \(V_A = 21.60\ \text{cm}^3\) and \(V_B = 25.00\ \text{cm}^3\):
\[ C_B = \frac{C_A V_A}{V_B} = \frac{0.100 \times 21.60}{25.00} = 0.0864\ \text{mol dm}^{-3} \]
(b)(iii) Mass of Na+ formed in solution
Moles of NaHSO4 reacted (= moles of Na2SO4 formed):
\[ n = \frac{0.100 \times 21.60}{1000} = 2.16 \times 10^{-3}\ \text{mol} \]
Each mole of Na2SO4 contains 2 moles of Na+, so:
\[ n(Na^+) = 2 \times 2.16 \times 10^{-3} = 4.32 \times 10^{-3}\ \text{mol} \]\[ \text{Mass of Na}^+ = 4.32 \times 10^{-3} \times 23.0 = 0.0994\ \text{g} \]
Reaction:
\[ NaHSO_4 + NaOH \to Na_2SO_4 + H_2O \quad (1:1) \]
(a) Titration results (indicator: methyl orange; pipette volume of B = 25.00 cm3)
| Burette readings (cm3) | Rough | 1st | 2nd |
|---|
| Final reading | 21.70 | 43.30 | 31.90 |
| Initial reading | 0.00 | 21.70 | 10.30 |
| Volume of A used | 21.70 | 21.60 | 21.60 |
Average volume of A (from concordant titres):
\[ V_A = \frac{21.60 + 21.60}{2} = 21.60\ \text{cm}^3 \]
(b)(i) Concentration of A in mol dm-3
Molar mass of NaHSO4 \(= 23.0 + 1.00 + 32.0 + (4 \times 16.0) = 120.0\ \text{g mol}^{-1}\).
\[ [A] = \frac{12.0}{120.0} = 0.100\ \text{mol dm}^{-3} \]
(b)(ii) Concentration of B in mol dm-3
Since the reaction is 1:1, \(\dfrac{C_A V_A}{C_B V_B} = 1\), with \(V_A = 21.60\ \text{cm}^3\) and \(V_B = 25.00\ \text{cm}^3\):
\[ C_B = \frac{C_A V_A}{V_B} = \frac{0.100 \times 21.60}{25.00} = 0.0864\ \text{mol dm}^{-3} \]
(b)(iii) Mass of Na+ formed in solution
Moles of NaHSO4 reacted (= moles of Na2SO4 formed):
\[ n = \frac{0.100 \times 21.60}{1000} = 2.16 \times 10^{-3}\ \text{mol} \]
Each mole of Na2SO4 contains 2 moles of Na+, so:
\[ n(Na^+) = 2 \times 2.16 \times 10^{-3} = 4.32 \times 10^{-3}\ \text{mol} \]\[ \text{Mass of Na}^+ = 4.32 \times 10^{-3} \times 23.0 = 0.0994\ \text{g} \]
