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 moldm\(^3\) of HCl. C is a solution containing 14.3g of Na2CO\(_{3}\).xH\(_2\)O in 500 cm\(^3\) of solution.
a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0cm\(^3\) 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;
Na\(_2\)CO\(_3\) . xH\(_2\)O + 2HCl\(_{(aq)}\) \(\to\) 2NaCl\(_{(aq)}\) + CO\(_{2(g)}\) + (x + 1)\(_3\)H\(_2\)O\(_{(l)}\)
(b) From your results and the information provided. calculate the:
(i) concentration of C in moldm\(^{-3}\)
(ii) concentration of C in gdm\(^{-3}\)
(iii) molar mass of Na\(_2\)CO\(_3\) . xH\(_2\)O
(iv) the value of x in Na\(_2\)CO\(_3\) . xH\(_2\)O. [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.
Explanation
Indicator = Methyl Orange
Volume of the base used = 25.00cm\(^3\)
| Titration | Rough Titre | 1st Titre | 2nd Titre | 3rd Titre |
| Final burette readings cm\(^3\) | 24.70 | 24.80 | 24.70 | 24.90 |
| Initial burette reading cm\(^3\) | 0.00 | 0.00 | 0.00 | 0.00 |
| Volume of acid used cm\(^3\) | 24.70 | 24.80 | 24.70 | 24.90 |
Average Titre = \(\frac{1st + 2nd + 3rd}{3}\)
= \(\frac{24.80 + 24.70 + 24.90}{3}\)
= 24.80cm\(^3\)
Alternatively 2 concordant titres can be used to calculate average titre
Equation of thereaction; Na\(_2\)CO\(_3\).XH\(_2\)O + 2HCl\(_{(aq)}\) \(\to\) 2Nacl\(_{(aq)}\) \(\to\) 2Nacl\(_{(aq)}\) + CO\(_{2(aq)}\) + (X + 1)H\(_2\)O\(_{(l)}\)
\(\frac{C_AV_A}{C_BV_B} = \frac{n_A}{n_B}\)
C\(_A\) = Molar concentration of HCl\(_{(aq)}\) in moldm\(^{3}\)
V\(_A\) = Volume of acid used in cm\(^3\) = 24.80cm\(^3\)
C\(_B\) = Molar concentration of Na\(_2\)Cu\(_3\) . xH\(_2\)O in molddm\(^3\)
n\(_A\) = 2
n\(_B\) = 1
(b)(i) Concentration of C in moldm\(^{-3}\) = ?
From the equation reaction;
\(\frac{C_AV_A}{C_BV_B} = \frac{n_A}{n_B}\)
C\(_A\) = 0.200 molddm\(^{-3}\)
C\(_B\) = ?
V\(_A\) = 24.80 cm
V\(_B\) = 25.00 cm\(^3\)
Substitution of known values
\(\frac{0.200 \times 24.80}{C_B \times 25.00} = \frac{2}{1}\)
C\(_B\) = \(\frac{1 \times 0.200 \times 24.80}{2 \times 25.00}\)
C\(_B\) = 0.0992 moldm\(^{-3}\)
(ii) Concentration of C in gdm\(^3\) = ?
500 cm\(^2\) \(\to\) 14.3 g
1000cm\(^3\) \(\to\) \(\frac{14.3}{500}\) x 1000g
= 28.6 dm\(^{-3}\)
(iii) Molar mass of Na\(_2\)CO\(_3\) . XH\(_2\)O
Molar conc in moldm\(^{-3}\) = \(\frac{\text{conc. in gdm}^{-3}} {\text{molar mass} }\)
Molar mass gmol\(^{-1}\) = \(\frac{\text{conc in g dm}^{-3}}{\text{molar conc.in moldm}^{-3}}\)
\(\frac{28.6 gdm^{-3}}{0.0992 moldm^{-3}}\)
= 288.3065
= 288 gmol\(^{-1}\)
(iv) Value of x in Na\(_2\)CO\(_3\) . xH\(_2\)O?
[H = 1.0, C = 12.0, O = 16.0, Na = 23.0]
N\(_2\)CO\(_3\) . xH\(_2\)O = 288
2(23) + 12 + 3 (16) + x(2(1) + 16) = 288
46 + 12 +48 +18x = 288
106 + 18x = 288
18x = 288 - 106
18x = 182
x = \(\frac{182}{18}\)
x = 10.11
x = 10