Burette readings (initial and final) must be given to two decimal places. Volume or pipete 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 containing 12.0g dm\(^{-3}\) NaHSO\(_4\) NaHSO\(_4\) P is a solution containing NaOH
(a) Put A into the burette and titrate it against 20.0cm\(^3\) or 25.0m\(^3\) portions of B using methyl orange as an indicator. Repeat the titration to obtain consistent titres. Tabulate your readings and calculate the average volume A used. The equation for the reaction involved in the titration is ;
NaHSO\(_{4(aq)}\) + NaOH\(_{(aq)}\) \(\to\) Na\(_2\)SO\(_{4(aq)}\) + H\(_2\)O\(_{(l)}\)
[H 1.00, O = 16.0; Na = 23.0, S = 32.0]
(b) From your results and the information provided above calculate the:
(i) concentration of A in mol dm\(^{-3}\)
(ii) concentration of B in mol dm\(^{-3}\)
(iii) mass of Na\(^+\) formed in solution during the titration.
Explanation
(a) Volume of base used (pipette) = 25.00 cm\(^3\); indicator - methyl orange
| Burette reading cm\(^3\) | Rough | 1st Titration | 2nd Titration |
| Final burette readings (cm\(^3\)) | 21.70 | 43.30 | 31.90 |
| Initial burette readings (cm\(^3\)) | 0.00 | 21.60 | 10.30 |
| Volume of acid used (cm\(^3\)) | 21.70 | 21.60 | 21.60 |
Average volume of acid = \(\frac{21.60 + 21.60}{2}\)
V\(_A\) = 21.60cm\(^3\)
(b)(i) Concentration of A in mol dm\(^{-3}\)
Molar mass of NaHSO\(_4\) = 23 + 1 + 32 + (4 x 16) = 120g mol\(^{-1}\)
Concentration of A I mol/dm3 = \(\frac{ mass concentration of A}{Molar mass of A}\)
= \(\frac{12.0}{120}\) = 0.100mol\(dm^3\)
(ii) concentration of B in mol dm\(^{-3}\)
\(\frac{C_AV_A}{C_BV_B}\) = 1
C\(_B\) = \(\frac{C_AV_A}{V_B}\)
= \(\frac{0.10 \times 21.6}{25}\)
= 0.0864 mol dm\(^{-3}\) (3 s.fig)
iii) Mass of \(Na^+\) formed in solution . Amount of NaOH in the solution = \(\frac{25 \times0.0864}{1000}\) = 0.00216 moles
Mole ratio NaOH : \(Na_{2}SO_{4}\) = 1 : 1 Amount of \(Na_{2}SO_{4}\) = 0.00216 moles. Atomic mass of \(Na^+\) = 23
Amount of \(Na^+\) = 2 x 0.00216. Mass of \(Na^+\) formed = 23 x 2 x 0.00216 = 0.9936g. ⇒ 0.10g