TEST OF PRACTICAL KNOWLEDGE QUESTION
All your burette readings (initials and final) as well as the size of your pipette must be recorded but no account experimental procedure is required. All calculations must be done in your answer booklet.
A is \(0.100\ \text{mol dm}^{-3}\) HNO\(_3\). B is a solution containing \(2.50\ \text{g}\) of a mixture of Na\(_2\)CO\(_3\) and Na\(_2\)SO\(_4\) in \(250\ \text{cm}^3\) of solution.
(a) Put into the burette and titrate it against \(20.0\ \text{cm}^3\) or \(25.0\text{cm}^3\) portions of B using methyl orange as indicator. Repeat the exercise to obtain consistent titre values. Tabulate your readings and calculate the average volume of acid A used. The equation of reaction is
\[
\text{Na}_2\text{CO}_{3(aq)} + 2\text{HNO}_{3(aq)} \to \text{NaNO}_{3(aq)} + \text{H}_2\text{O}_{(l)} + \text{CO}_{2(g)}
\]
(b) From your results and the information provided, calculate the:
(i) concentration of B in moldm\(^{-3}\);
(ii) concentration of Na\(_2\)CO\(_3\) in B in gdm\(^{-3}\);
(iii) percentage of Na\(_2\)CO\(_3\), in the mixture. [Na\(_2\)CO\(_3\) = 106]
(a) The average titre is read from the candidate's own concordant burette readings. Only the Na2CO3 in B reacts with the acid (Na2SO4 is inert to HNO3). Taking a typical average titre of A as 20.00 cm3 for a 25.0 cm3 portion of B illustrates the working.
(b)(i) Concentration of Na2CO3 in B (mol dm-3)
Amount of HNO3 \(= 0.100 \times \dfrac{20.00}{1000} = 2.00\times10^{-3}\ \text{mol}\).
From \(Na_2CO_3 + 2HNO_3 \rightarrow 2NaNO_3 + H_2O + CO_2\), amount of Na2CO3 \(= \dfrac{2.00\times10^{-3}}{2} = 1.00\times10^{-3}\ \text{mol}\) in 25.0 cm3.
Concentration \(= 1.00\times10^{-3} \times \dfrac{1000}{25.0} = \mathbf{0.0400\ mol\ dm^{-3}}\).
(ii) Concentration of Na2CO3 in g dm-3
\(= 0.0400 \times 106 = \mathbf{4.24\ g\ dm^{-3}}\).
(iii) Percentage of Na2CO3 in the mixture
Total B contains 2.50 g in 250 cm3, i.e. \(2.50 \times \dfrac{1000}{250} = 10.0\ \text{g dm}^{-3}\).
\(\%\,Na_2CO_3 = \dfrac{4.24}{10.0} \times 100 = \mathbf{42.4\%}\).
(a) The average titre is read from the candidate's own concordant burette readings. Only the Na2CO3 in B reacts with the acid (Na2SO4 is inert to HNO3). Taking a typical average titre of A as 20.00 cm3 for a 25.0 cm3 portion of B illustrates the working.
(b)(i) Concentration of Na2CO3 in B (mol dm-3)
Amount of HNO3 \(= 0.100 \times \dfrac{20.00}{1000} = 2.00\times10^{-3}\ \text{mol}\).
From \(Na_2CO_3 + 2HNO_3 \rightarrow 2NaNO_3 + H_2O + CO_2\), amount of Na2CO3 \(= \dfrac{2.00\times10^{-3}}{2} = 1.00\times10^{-3}\ \text{mol}\) in 25.0 cm3.
Concentration \(= 1.00\times10^{-3} \times \dfrac{1000}{25.0} = \mathbf{0.0400\ mol\ dm^{-3}}\).
(ii) Concentration of Na2CO3 in g dm-3
\(= 0.0400 \times 106 = \mathbf{4.24\ g\ dm^{-3}}\).
(iii) Percentage of Na2CO3 in the mixture
Total B contains 2.50 g in 250 cm3, i.e. \(2.50 \times \dfrac{1000}{250} = 10.0\ \text{g dm}^{-3}\).
\(\%\,Na_2CO_3 = \dfrac{4.24}{10.0} \times 100 = \mathbf{42.4\%}\).
