All your burette readings (initial and final), as well as the size of your pipette, must be recorded but on no account of experiment procedure is required. All calculations must be done in your answer book.
A is mol dm HCI. B is a solution containing 15.0 g dm of a mixture of NaCl and KHCO\(_3\).
(a) Put A burette and titrate it against 20.0cm\(^3\) or 25.0cm\(^3\) portions 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 involved in the titration is: HCl\(_{aq}\) + KHCO\(_{3(aq)}\) \(\to\) KCl\(_{(aq)}\) +CO\(_{2(g)}\)
(b) From your results and the information provided above, calculate the:
(i) concentration of KHCO\(_3\), in mol dm\(^{-3}\) in B;
(ii) mass of KHCO\(_3\), in g dm\(^{-3}\) in B
(ii) Percentage by mass of KHCO\(_{3}\) in the mixture, [H=1; C = 12; O = 16; K = 39]
(iv) mass of NaCl in the mixture.
Explanation
| Rough | 1st Reading | 2nd Reading | 3rd Reading | 3ed Reading |
| Final volume | 27.00 | 36.00 | 26.05 | 40.95 |
| Initial volume | 0.00 | 10.00 | 0.00 | 15.00 |
| Volume A used | 27.00 | 26.00 | 26.05 | 25.95 |
Volume pipette = 25cm\(^3\)
Volume of A used = \(\frac{26.00 + 26.95 + 25.95}{3}\)
= \(\frac{78.00}{3}\)
= 26.00 cm\(^3\)
Equation for reaction; HCL + KHCO\(_3\) \(\to\) KCl + CO\(_2\)
(i) Let C\(_A\) represent the conc. of A = 0.100 mol sm\(^{-3}\)
V\(_A\) volume of A = 26.00 cm\(^3\)
C\(_B\) conc. of B = ?
V\(_B\) volume of B = 25.00 cm\(^3\)
\(_n\)A mole of A = 1
\(_n\)B mole of B = 1
from \(\frac{C_A \times V_B}{C_B \times V_B} = \frac{1}{1}\)
= \(\frac{0.100 \times 26}{C_B \times 25} = \frac{1}{1}\)
C\(_B\) = \(\frac{0.100 \times 26}{25}\) = 0.104 mol dm\(^{-3}\)
(ii) Mass KHCO\(_3\) in mol/dm\(^3\)
Molar mass of KHCO\(_3\) = 39 + 1 + 12 + 16 x 3 = 100
Mass = 0.104 x 100 = 10.4g/dm\(^3\)
(iii) Percentage by mass of KHCO\(_3\) in the mixture
= \(\frac{15.0 - 10.4}{15} \times \frac{100}{1}\) % = \(\frac{4.6}{15} \times \frac{100}{1}\)% = 30.7%
(iv) Mass of NaCl in the mixture = 15.0 - 10.4
= 4.6g