Burette readings (initial and final) must be given to two decimal places. Volume of pipette 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 of hydrochloric acid. B is a solution containing 2.45g of anhydrous sodium trioxocarbonate (IV) in 250g of solution.
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^3\) portion of B using methyl orange as an indicator. Repeat the exercise to obtain consistent titres. Tabulate your burette reading and calculate the average volume of A used. The equation for the reaction involved in the titration is Na\(_2\)CO\(_{3(aq)}\) + 2HCI\(_{(aq)}\) + H\(_2\)O\(_{(l)}\)
(b) From your results and the information provided, calculate the:
(i) concentration of B in moldm\(^{-3}\)
(ii) concentration of A in moldm\(^{-3}\)
(iii) concentration of A in gdm\(^{-3}\)
(iv) volume of the gas evolved in the reaction at s.t.p.
[H = 1.00; C: 12.0; O = 16.0; Na = 23.0; Na = 23.0; Cl = 35.5; Molar Volume = 22.4 dm\(^3\)mol\(^{-3}\)]
(a)(i)
| Burette Reading (cm\(^3\) | Rough Titration (cm\(^3\) | 1st 2nd Titration (cm\(^3\) | 2nd Titration |
|
Final Burette Reading (cm\(^3\)) Initial Burette Reading (cm\(^3\)) Vol. of Acid Used (cm\(^3\)) |
21.00 4.00 17.00 |
38.50 21.00 17.50 |
23.30 5.70 17.60 |
Average Titre = \(\frac{17.50 + 17.60}{2}\)
= 17.55 cm\(^3\)
(b)(i) Na\(_2\)CO\(_{3(aq)}\) + 2HCl\(_{(aq)}\) \(\to\) 2NaCl\(_{(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)
Molar mass (Na\(_2\)CO\(_3\)) = (2 x 23) + 12 + (3 x 15) = 106 gmol\(^{-1}\)
250cm\(^3\) of solution contained 2.45g of Na\(_2\)CO\(_3\)
1000cm\(^3\) of solution contained = \(\frac{2.45 \times 1000}{250}\)
= 2.45 x 4
= 9.80g
Con. B (Na\(_2\)CO\(_3\)) = \(\frac{9.80}{106}\)
= 0.0925 mol dm\(^{-3}\)
(ii) Conc. of A in mol dm\(^{-3}\)
\(\frac{C_AV_A}{C_BV_B} = \frac{N_A}{N_B}\)
\(\frac{C_A \times 17.55}{0.0925 \times 25}\) = \(\frac{2}{1}\)
C\(_A = \frac{0.0925 \times 25 \times 2}{17.55}\)
= 0.264 mol dm\(^{-3}\)
(iii) Conc. of A in gdm\(^{-1}\)
Molar mass HCl = 1 + 35.5 = 36.5 g mol\(^{1}\)
Conc of A(gdm\(^{-3}\)) = 0.264 x 36.5
= 9.64 gdm\(^{-1}\)
(iv) volume of the gas evolved
1 mole of Na\(_2\)CO\(_3\) = 1 mole of CO\(_2\) at s.t.p
n(CO\(_2\)) = n(NaCO\(_3\)) = 0.092.5 mole
vol. of CO\(_2\) = n x volume
= 0.0925 x 22.4 dm\(^{-3}\)
= 2.07dm\(^3\) at s.t.p
Volume of CO\(_2\) liberated in 23 cm\(^3\) of Na\(_2\)O\(_3\) will be \(\frac{2.07 \times 25}{1000}\)
= 0.0518 dm\(^3\) or 51.8 cm\(^3\) at s.t.p
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}