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 solution containing 6.22 g of an acid H\(_2\)Y per dm\(^3\)
B contains 3.90 g of NaOH per dm\(^3\) of solution.
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^3\) portions of B using methyl orange as indicator. Repeat the titration to obtain consistent titres. Tabulate your burette readings and calculate the average volume of acid A used. The equation for the reaction involved in the titration is:
H\(_2\)\(_{(aq)}\) + 2NaOH\(_({aq})\) \(\to\) NaY\(_{(aq)}\) + 2H\(_2\)O\(_{(l)}\)
[H = 1.00; O = 16.0; Na = 23.0]
(b) From your results and the information provided above, calculate the:
(i) The concentration of B in moldm\(^{-3}\)
(ii) concentration of A in moldm\(^3\)
(ii) molar mass of H\(_2\)Y.
(c) State whether the pH of each of the following solutions is lower than 7, greater than 7 or equal to 7. The
(I) solution A before titration
(ii) solution B before titration
| ROUGH (cm\(^3\)) | 1st Titration (cm\(^3\)) | 2nd Titration (cm\(^3\) | |
| Initial Burette Reading | 27.00 |
49.50 |
36.00 |
| Initial Burette Reading | 01.00 | 26.00 | 12.40 |
| Vol. of Acid Used(cm\(^3\)) | 26.00 | 23.50 | 23.60 |
Average Titre = \(\frac{23.50 + 23.60}{2}\) = 23.55cm\(^{3}\)
(b)(i) Concentration of B in moldm\(^{-3}\)
Molar mass of NaOH = 23 + 16 + 1 = 40g mol\(^{-1})
concentration of B = \(\frac{3.90}{40}\)
= 0.0975 mol dm\(^{-1}\)
(ii) concentration of BAin mol dm\(^{-3}\)
\(\frac{C_AV_A}{C_BV_B}\) = \(\frac{1}{2}\)
C\(_A = \frac{C_BV_B}{2V_A}\)
C\(_A\) = \(\frac{0.975 \times 25}{2 \times 23.55}\)
= 0.0518 mol dm\(^{-3}\)
(ii) Molar mass of H\(_2\)Y
Molar mas of H\(_2\)Y = \(\frac{6.22}{0.0518}\)
= 120 gmol\(^{-1}\)
(c)(i) Less than 7
(ii) Greater than 7
[C = (NH\(_4\))\(_2\)SO\(_{4(aq)}\); D = FeCl\(_{3(aq)}\)
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