All your burette readings (initial 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 book.
A is a solution of HCI containing 5.0g dm\(^{-3}\). B is a solution of impure KOH containing 6.50g dm\(^{-3}\).
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 Tabulate your burette readings and calculate the average volume of A used. The equation for the reaction involved in the titration is: HCI\(_{(aq)}\) KOH\(_{(aq)}\) \(\to\) Cl\(_{(aq)}\) H\(_2\)O\(_{(l)}\)
(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) percentage purity of KOH in B [H= 1; CI = 35.5; KOH = 56.0g mol\(^{-1}\)]
Explanation
| Titration number | Rough | 1 | 2 | 3 |
| Final reading | 17.90 | 18.00 | 18.00 | 28.00 |
| Initial reading |
0.00 |
0.00 | 0.00 | 0.00 |
| Volume of solution A used/cm\(^3\) | 17.90 | 18.00 | 18.00 | 18.00 |
HCl\(_{(aq)}\) + KOH\(_{(aq)}\) + H\(_2\)O\(_{(l)}\) Average volume of acid used (cm\(^{3}\)
= \(\frac{18.00 + 18.00 + 18.00}{3} = \frac{54.00}{3}\) = 18.00 cm\(^3\)
(b)(i) To calculate the concentration of solution A molar mass of HCl = 1 + 35.5 = 36.5 mol \(^{-1}\)
concentration in g/dm\(^3\) = 5
concentration A (C\(_A\)) = \(\frac{5.0}{36.5}\) = 0.137 mol/dm\(^3\)
(ii) To calculate the concentration of solution B \(\frac{C_AV_A}{C_BV_B}\) = \(\frac{N_A}{N_B}\)
HCl\(_{(aq)}\) + KOH\(_{(aq)}\) \(\to\) KCl\(_{(aq)}\) + H\(_2\)O\(_{(l)}\)
N\(_A\) = mole ratio of acid = 1
N\(_B\) = mole ration of base = 1
C\(_A\) = concentration acid = 0.137 mol/dm\(^3\)
C\(_B\) = concentration of base = ?
V\(_A\) = volume of acid used in cm\(^3\) = 18.00 cm\(^3\)
V\(_B\) = volume of base used = 25.00 cm\(^{3}\)
C\(_B\) = \(\frac{C_AV_A \times N_B}{V_B \times N_A} = \frac{0.137 \times 18 \times 1}{25 \times 1}\)
= \(\frac{2.466}{25}\)
= 0.099 mol/dm\(^{3}\)
(iii) To calculate percentage purity of KOH in B; Molar mass of KOH = 56g/dm\(^3\)
Concentration in mol/dm\(^3\) = \(\frac{\text{concentration in g/dm}^3}{\text{molar mass}}\)
Molar mass = 0.099 x 56
= 5.54 g/dm\(^3\)
Percentage purity = \(\frac{\text{mass concentrate}}{\text{mass of impure KOH}}\) x \(\frac{100}{1}\)
= \(\frac{5.54}{6.50} \times \frac{100}{1}\)
= 85.23%