TEST OF PRACTICAL KNOWLEDGE QUESTION
Burette reading (initial and final) must be given to two decimal places. Volume of pipette used must be recorded but no account of experimental procedure is required. All calculations must be done in you answer book.
A is a solution containing 6.3 g dm\(^{-3}\) of HNO\(_3\), B is a solution Na\(_2\)CO\(_3\)
(a) Put A Into the burette and titrate it against 20.0 cm\(^3\) portions of B using menyl Indicator. Record the volume of your pipette. Repeat the titration to obtain consistent titres. Tabulate your burette readings and calculate the average volume of A used. The equation for the reaction involved in the titration is:
2HNO\(_{3(aq)}\) + Na\(_2\)CO\(_{3(aq)}\) \(\to\) 2NaNO\(_{3(aq)}\) + CO\(_{2(g)}\)
(b) From your results and information provided above, calculate the;
(i) concentration of B in mol dm\(^{-3}\)
(ii) concentration of B in g dm\(^{-3}\):
(iii) mass of sodium ions in 1:0 dm\(^{-3}\) of B
[H = 1; C = 1; O = 16; N = 14; Na = 23]
Titration of A (HNO3) against B (Na2CO3)
The concentration of A is fixed by the data: molar mass of HNO3 \( = 1+14+48 = 63\ \text{g mol}^{-1}\), so
\[ [A] = \frac{6.3}{63} = 0.10\ \text{mol dm}^{-3} \]
The actual average titre depends on the candidate's burette readings. Taking a representative average titre of \(V_A = 25.00\ \text{cm}^3\) of A for 20.0 cm3 of B, the working is:
Equation: \[ 2HNO_3 + Na_2CO_3 \to 2NaNO_3 + CO_2 + H_2O \]
(i) Concentration of B in mol dm-3
Moles of HNO3 used \( = 0.10 \times \dfrac{25.00}{1000} = 2.5 \times 10^{-3}\ \text{mol} \)
Mole ratio HNO3 : Na2CO3 = 2 : 1, so moles of Na2CO3 \( = \dfrac{2.5\times10^{-3}}{2} = 1.25\times10^{-3}\ \text{mol} \) in 20.0 cm3.
\[ [B] = \frac{1.25\times10^{-3}}{20.0/1000} = \mathbf{0.0625\ mol\,dm^{-3}} \]
(ii) Concentration of B in g dm-3
Molar mass of Na2CO3 \( = (2\times23)+12+(3\times16) = 106\ \text{g mol}^{-1} \)
\[ = 0.0625 \times 106 = \mathbf{6.63\ g\,dm^{-3}} \]
(iii) Mass of sodium ions in 1.0 dm3 of B
Each Na2CO3 gives 2 Na+, so moles of Na+ \( = 2 \times 0.0625 = 0.125\ \text{mol} \)
\[ \text{Mass of Na}^+ = 0.125 \times 23 = \mathbf{2.88\ g} \]
(The numerical answers scale with the candidate's actual average titre; the method is exactly as shown.)
Titration of A (HNO3) against B (Na2CO3)
The concentration of A is fixed by the data: molar mass of HNO3 \( = 1+14+48 = 63\ \text{g mol}^{-1}\), so
\[ [A] = \frac{6.3}{63} = 0.10\ \text{mol dm}^{-3} \]
The actual average titre depends on the candidate's burette readings. Taking a representative average titre of \(V_A = 25.00\ \text{cm}^3\) of A for 20.0 cm3 of B, the working is:
Equation: \[ 2HNO_3 + Na_2CO_3 \to 2NaNO_3 + CO_2 + H_2O \]
(i) Concentration of B in mol dm-3
Moles of HNO3 used \( = 0.10 \times \dfrac{25.00}{1000} = 2.5 \times 10^{-3}\ \text{mol} \)
Mole ratio HNO3 : Na2CO3 = 2 : 1, so moles of Na2CO3 \( = \dfrac{2.5\times10^{-3}}{2} = 1.25\times10^{-3}\ \text{mol} \) in 20.0 cm3.
\[ [B] = \frac{1.25\times10^{-3}}{20.0/1000} = \mathbf{0.0625\ mol\,dm^{-3}} \]
(ii) Concentration of B in g dm-3
Molar mass of Na2CO3 \( = (2\times23)+12+(3\times16) = 106\ \text{g mol}^{-1} \)
\[ = 0.0625 \times 106 = \mathbf{6.63\ g\,dm^{-3}} \]
(iii) Mass of sodium ions in 1.0 dm3 of B
Each Na2CO3 gives 2 Na+, so moles of Na+ \( = 2 \times 0.0625 = 0.125\ \text{mol} \)
\[ \text{Mass of Na}^+ = 0.125 \times 23 = \mathbf{2.88\ g} \]
(The numerical answers scale with the candidate's actual average titre; the method is exactly as shown.)
