All your burette readings (initial and final), as well as the size of your pipette, must be recorded but on no account of experimental procedure is required. All calculations must be done in your answer book.
(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 indicator. Record the volume of your pipette. Tabulate your burette readings and calculate the average volume of A used. The equation for the reaction is 2NaOH\(_{(aq)}\) + H\(_2\)X\(_{(aq)}\) \(\to\) 2H\(_2\)O\(_{(l)}\)
(b) From your results and the information provided, calculate the;
(i) Concentration of solution B in mol dm\(^{-3}\)
(ii) concentration of solution A in mol dm\(^{-3}\)
(iii) molar mass of the acid H\(_2\)X. [H = 1, O = 16, Na = 23]
| Rough | 1st Reading | 2nd Reading | 3rd Reading | |
| Final Burette Reading | 24.0 | 23.60 | 24.15 | 28.55 |
| Initial Burette Reading | 0.00 | 0.00 | 0.50 | 5.00 |
| Volume of A used | 24.00 | 23.60 | 23.65 | 23.55 |
Indicator used - Methyl orange
Colour change at end point from yellow to pink
Volume of pipette 25 cm\(^3\)
Average titre = \(\frac{23.60 + 23.65 + 23.55}{3}\) = \(\frac{70.80}{3}\)
= 23.60 cm\(^3\)
(b)(i) Concentration of B Molar (formula) mass of NaOH = (23 + 16 + 1) or 40g mol\({-1}\)
Mass of NaOH in 1 dm\(^3\) of solution of B = \(\frac{2.15 \times 1000}{250}\)
OR 2.15 x 4 = 8.60 g dm\(^{-3}\)
Concentration of solution of B = \(\frac{8.60}{40}\) = 0.215 mol dm\(^{-3}\)
(ii) Equation for the reaction; 2 NaOH + H\(_2\)X \(\to\) Na\(_2\)X + 2H\(_2\)O
\(\frac{\text{No. of moles of A(C\(_A\)) x volume of A (V\(_A\))}}{\text{No of moles of B(C\(_{B}\)) x volume of B(V\(_B\))}}\) = molar ratio of \(\frac{A}{B}\)
\(\frac{C_A \times V_A}{C_B \times V_B} = \frac{1}{2}\)
\(\frac{C_A \times 23.6}{0.125 \times 25} = \frac{1}{2}\)
C\(_A\) = \(\frac{0.245 \times 25}{2 \times 23.6}\) = 0.144 mol dm\(^{-3}\) to 3 sig. fig.
(iii) Molar mass of H\(_2\)X, concentration in mol dm\(^{3}\) = \(\frac{\text{mass per dm\(^3\)}}{\text{Molar mass}}\)
Molar mass of H\(_2\)X = \(\frac{3.70}{0.114}\)
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