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 a solution containing 15.8 g dm\(^3\) of Na\(_2\)S\(_2\)O\(_3\). B was obtained by dissolving 9.0 g of an impure sample of I\(_2\) in aqueous Kl and the solution made up to 1 dm\(^3\).
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^3\) portions of B. Use starch solution as indicator. Repeat the titration to obtain concordant titre values. Tabulate your results and calculate the average volume of A used. The equation for the reaction involved in the titration is I\(_2\) + 2S\(_2\)O\(_3\) \(\to\) 2I\(^-\) + S\(_4\)O\(_6^{2-}\).
(b) From your results and the information provided, calculate the:
(i) concentration of A in mol dm\(^{-3}\)
(ii) concentration of I\(_2\) in B in mol dm\(^{-3}\);
(iii) percentage by mass of I\(_2\) in the sample
(c) Give reasons why the starch indicator was not added to the titration mixture at the beginning of the titration. [O = 16.0, Na = 23.0, S = 32.0, 1 = 127.0] Credit will be given for strict adherence to the instructions for observations precisely recorded and for accurate inferences. AIl tests, observations and inferences must be clearly entered in the booklet in ink at the time they are made.
(a)
| Rough Tree | 1st Titre | 2nd Titre | 3rd Titre | |
| Final reading cm\(^3\) | 12.60 | 25.10 | 12.40 | 25.00 |
| Initial reading cm\(^3\) | 0.60 | 12.60 | 0.00 | 12.40 |
| Volume of acid cm\(^3\) | 12.60 | 12.50 | 12.40 | 12.60 |
Average Titre = \(\frac{12.50 + 12.40 + 12.60}{3}\) = 12.50cm\(^3\)
Equation of the reaction I\(_2\) + 2S\(_2\)O\(_3^{2-}\) \(\to\) 2I\(^-\) + S\(_4\)O\(_6^{2-}\)
(b)(i) Concentration of A in moldm\(^{-3}\)
= \(\frac{\text{Concentration of A in gdm}^{-3}}{\text{Molar mass of A in g mol}^{-1}}\)
= \(\frac{15.8gdm^{-3}}{\text{Molar mass of A}}\)
Molar mass Na\(_2\)S\(_2\)O\(_3\) = (23)2 + (32) 2 + (16)3
= 46 + 64 + 48
= 158 g mol\(^{-1}\)
concentration of A in moldm\(^{-3}\)
= \(\frac{15.8gdm^{-3}}{158gmol^{-1}}\) = 0.100 moldm\(^{-3}\)
(ii) Concentration of I\(_2\) in B moldm\(^{-3}\)
\(\frac{C_AV_A}{C_BV_B} = \frac{n_A}{n_B}\)
Where;
C\(_A\) = 0.100 moldm\(^{-3}\)
V\(_A\) = 12.50 cm\(^2\)
C\(_B\) = ?
V\(_B\) = 25.00 cm\(^2\)
C\(_B\) = \(\frac{0.100 \times 12.50 \times 1}{2 \times 25.00}\)
C\(_B\) = \(\frac{1.25}{50}\)
C\(_B\)= 0.0250 moldm\(^{-3}\)
(iii) percentage by ass of I\(_2\) in the sample
= \(\frac{\text{Concentration in gdm}^{-3}}{9.0 gdm^{-3} \text{of impure}} \times \frac{100%}{1}\)
Concentration in gdm\(^{-3}\) = \(\frac{\text{Concentration in gdm}^{-3} Pure}{ \text{ Molar mass of } I_2}\)
0.0250 gdm\(^{-3}\) = \(\frac{\text{Concentration in gdm}^{-3} Pure}{ \text{Molar mass of } I_2}\)
I\(_2\) = 127 x 2 = 254/mol
0.0250 gdm\(^{-3}\) = \(\frac{\text{Concentration in gdm}^{-3} (Pure)}{ \text{Molar mass of }I_2}\)
Concentration in gdm\(^{-3}\) (Pure) = 0.0250 x 254
= 6.35 gdm\(^{-3}\)
Percentage by mass of I\(_2\) in the sample
= \(\frac{6.35}{9.0} \times \frac{100%}{1}\) = 70.555
= 70.56%
(c) Starch indicator was not added to the titration mixture at the beginning of the titration in order to obtain an accurate end-point or to prevent the formation of the complex which reduces the accuracy of the titre value.
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