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 of experimental procedure is required. All calculations must be done in your answer booklet.
A is \(0.200\ \mathrm{moldm^3}\) of HCl. C is a solution containing \(14.3\mathrm{g}\) of \(\mathrm{Na_2CO_3\ .\ xH_2O}\) in \(500\ \mathrm{cm^3}\) of solution.
a) Put A into the burette and titrate it against \(20.0\ \mathrm{cm^3}\) or \(25.0\mathrm{cm^3}\) portions of C using methyl orange as indicator. Repeat the titration to obtain Consistent titre values. Tabulate vour results and calculate the average volume of A used. The equation for the reaction is;
\[
\mathrm{Na_2CO_3\ .\ xH_2O + 2HCl_{(aq)} \to 2NaCl_{(aq)} + CO_{2(g)} + (x + 1)_3H_2O_{(l)}}
\]
(b) From your results and the information provided. calculate the:
(i) concentration of C in \(\mathrm{moldm^{-3}}\)
(ii) concentration of C in \(\mathrm{gdm^{-3}}\)
(iii) molar mass of \(\mathrm{Na_2CO_3\ .\ xH_2O}\)
(iv) the value of x in \(\mathrm{Na_2CO_3\ .\ xH_2O}\). [H =1.0; C = 12.0; O = 16.0; Na = 23.0]
Credit will be given for strict adherence to the instruction, for observations precisely recorded and for accurale references. All tests. obsenations and influences must be cleary entered in the booklet in ink at the same time they are made.
Indicator: methyl orange. Volume of C (base) pipetted: 25.00 cm3.
(a) Burette readings and average titre
| Titration | Rough | 1st | 2nd | 3rd |
|---|
| Final burette reading / cm3 | 24.70 | 24.80 | 24.70 | 24.90 |
| Initial burette reading / cm3 | 0.00 | 0.00 | 0.00 | 0.00 |
| Volume of A used / cm3 | 24.70 | 24.80 | 24.70 | 24.90 |
Using the three consistent titres:
\[ \text{Average titre} = \frac{24.80 + 24.70 + 24.90}{3} = 24.80\ \text{cm}^3 \]
(b)(i) Concentration of C in mol dm-3
Equation: \(Na_2CO_3\cdot xH_2O + 2HCl \rightarrow 2NaCl + CO_2 + (x+1)H_2O\), so \(\dfrac{C_AV_A}{C_BV_B} = \dfrac{n_A}{n_B} = \dfrac{2}{1}\).
With \(C_A = 0.200\ \text{mol dm}^{-3}\), \(V_A = 24.80\ \text{cm}^3\), \(V_B = 25.00\ \text{cm}^3\):
\[ \frac{0.200 \times 24.80}{C_B \times 25.00} = \frac{2}{1} \]
\[ C_B = \frac{1 \times 0.200 \times 24.80}{2 \times 25.00} = 0.0992\ \text{mol dm}^{-3} \]
(ii) Concentration of C in g dm-3
C contains 14.3 g in 500 cm3, so
\[ \frac{14.3}{500} \times 1000 = 28.6\ \text{g dm}^{-3} \]
(iii) Molar mass of Na2CO3·xH2O
\[ M = \frac{\text{concentration in g dm}^{-3}}{\text{concentration in mol dm}^{-3}} = \frac{28.6}{0.0992} = 288\ \text{g mol}^{-1} \]
(iv) Value of x
\[ 2(23.0) + 12.0 + 3(16.0) + x(2(1.0)+16.0) = 288 \]
\[ 46 + 12 + 48 + 18x = 288 \]
\[ 106 + 18x = 288 \]
\[ 18x = 182,\qquad x = \frac{182}{18} = 10.11 \approx 10 \]
The salt is Na2CO3·10H2O.
Indicator: methyl orange. Volume of C (base) pipetted: 25.00 cm3.
(a) Burette readings and average titre
| Titration | Rough | 1st | 2nd | 3rd |
|---|
| Final burette reading / cm3 | 24.70 | 24.80 | 24.70 | 24.90 |
| Initial burette reading / cm3 | 0.00 | 0.00 | 0.00 | 0.00 |
| Volume of A used / cm3 | 24.70 | 24.80 | 24.70 | 24.90 |
Using the three consistent titres:
\[ \text{Average titre} = \frac{24.80 + 24.70 + 24.90}{3} = 24.80\ \text{cm}^3 \]
(b)(i) Concentration of C in mol dm-3
Equation: \(Na_2CO_3\cdot xH_2O + 2HCl \rightarrow 2NaCl + CO_2 + (x+1)H_2O\), so \(\dfrac{C_AV_A}{C_BV_B} = \dfrac{n_A}{n_B} = \dfrac{2}{1}\).
With \(C_A = 0.200\ \text{mol dm}^{-3}\), \(V_A = 24.80\ \text{cm}^3\), \(V_B = 25.00\ \text{cm}^3\):
\[ \frac{0.200 \times 24.80}{C_B \times 25.00} = \frac{2}{1} \]
\[ C_B = \frac{1 \times 0.200 \times 24.80}{2 \times 25.00} = 0.0992\ \text{mol dm}^{-3} \]
(ii) Concentration of C in g dm-3
C contains 14.3 g in 500 cm3, so
\[ \frac{14.3}{500} \times 1000 = 28.6\ \text{g dm}^{-3} \]
(iii) Molar mass of Na2CO3·xH2O
\[ M = \frac{\text{concentration in g dm}^{-3}}{\text{concentration in mol dm}^{-3}} = \frac{28.6}{0.0992} = 288\ \text{g mol}^{-1} \]
(iv) Value of x
\[ 2(23.0) + 12.0 + 3(16.0) + x(2(1.0)+16.0) = 288 \]
\[ 46 + 12 + 48 + 18x = 288 \]
\[ 106 + 18x = 288 \]
\[ 18x = 182,\qquad x = \frac{182}{18} = 10.11 \approx 10 \]
The salt is Na2CO3·10H2O.
