TEST OF PRACTICAL KNOWLEDGE QUESTION
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 \(20.0\text{cm}^3\) or \(25.0\text{cm}^3\) 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:
\[
\mathrm{HCl}_{aq} + \mathrm{KHCO}_{3(aq)} \to \mathrm{KCl}_{(aq)} + \mathrm{CO}_{2(g)}
\]
(b) From your results and the information provided above, calculate the:
(i) concentration of KHCO\(_3\), in \(\text{mol dm}^{-3}\) in B;
(ii) mass of KHCO\(_3\), in \(\text{g dm}^{-3}\) 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.
Titration of A (HCl) against B (NaCl + KHCO3 mixture)
| |
Rough |
1st Reading |
2nd Reading |
3rd Reading |
| Final burette reading (cm3) |
27.00 |
36.00 |
26.05 |
40.95 |
| Initial burette reading (cm3) |
0.00 |
10.00 |
0.00 |
15.00 |
| Volume of A used (cm3) |
27.00 |
26.00 |
26.05 |
25.95 |
Volume of pipette = 25.0 cm3
Average volume of A used:
\[
\frac{26.00+26.05+25.95}{3}
=\frac{78.00}{3}
=26.00\text{ cm}^3
\]
Equation for the reaction:
\[
\mathrm{HCl}_{(aq)}+\mathrm{KHCO}_{3(aq)}
\rightarrow \mathrm{KCl}_{(aq)}+\mathrm{CO}_{2(g)}+\mathrm{H_2O}_{(l)}
\]
(i) Concentration of KHCO3 in B
Concentration of A (HCl) = 0.100 mol dm-3
Volume of A used = 26.00 cm3
Volume of B pipetted = 25.0 cm3
Mole ratio, HCl : KHCO3 = 1 : 1
\[
\frac{C_A V_A}{C_B V_B}=\frac{1}{1}
\]
\[
\frac{0.100 \times 26.00}{C_B \times 25.0}=\frac{1}{1}
\]
\[
C_B=\frac{0.100\times26.00}{25.0}
=0.104\text{ mol dm}^{-3}
\]
Concentration of KHCO3 = 0.104 mol dm-3
(ii) Mass of KHCO3 in g dm-3
Molar mass of KHCO3:
\[
39+1+12+(16\times3)=100\text{ g mol}^{-1}
\]
\[
\text{Mass concentration of KHCO}_3
=0.104\times100
=10.4\text{ g dm}^{-3}
\]
Mass of KHCO3 = 10.4 g dm-3
(iii) Percentage by mass of KHCO3 in the mixture
Total mass of mixture = 15.0 g dm-3
\[
\%\text{ KHCO}_3
=\frac{10.4}{15.0}\times100
=69.3\%
\]
Percentage by mass of KHCO3 = 69.3%
(iv) Mass of NaCl in the mixture
\[
\text{Mass of NaCl}
=15.0-10.4
=4.6\text{ g dm}^{-3}
\]
Mass of NaCl = 4.6 g dm-3
Titration of A (HCl) against B (NaCl + KHCO3 mixture)
| |
Rough |
1st Reading |
2nd Reading |
3rd Reading |
| Final burette reading (cm3) |
27.00 |
36.00 |
26.05 |
40.95 |
| Initial burette reading (cm3) |
0.00 |
10.00 |
0.00 |
15.00 |
| Volume of A used (cm3) |
27.00 |
26.00 |
26.05 |
25.95 |
Volume of pipette = 25.0 cm3
Average volume of A used:
\[
\frac{26.00+26.05+25.95}{3}
=\frac{78.00}{3}
=26.00\text{ cm}^3
\]
Equation for the reaction:
\[
\mathrm{HCl}_{(aq)}+\mathrm{KHCO}_{3(aq)}
\rightarrow \mathrm{KCl}_{(aq)}+\mathrm{CO}_{2(g)}+\mathrm{H_2O}_{(l)}
\]
(i) Concentration of KHCO3 in B
Concentration of A (HCl) = 0.100 mol dm-3
Volume of A used = 26.00 cm3
Volume of B pipetted = 25.0 cm3
Mole ratio, HCl : KHCO3 = 1 : 1
\[
\frac{C_A V_A}{C_B V_B}=\frac{1}{1}
\]
\[
\frac{0.100 \times 26.00}{C_B \times 25.0}=\frac{1}{1}
\]
\[
C_B=\frac{0.100\times26.00}{25.0}
=0.104\text{ mol dm}^{-3}
\]
Concentration of KHCO3 = 0.104 mol dm-3
(ii) Mass of KHCO3 in g dm-3
Molar mass of KHCO3:
\[
39+1+12+(16\times3)=100\text{ g mol}^{-1}
\]
\[
\text{Mass concentration of KHCO}_3
=0.104\times100
=10.4\text{ g dm}^{-3}
\]
Mass of KHCO3 = 10.4 g dm-3
(iii) Percentage by mass of KHCO3 in the mixture
Total mass of mixture = 15.0 g dm-3
\[
\%\text{ KHCO}_3
=\frac{10.4}{15.0}\times100
=69.3\%
\]
Percentage by mass of KHCO3 = 69.3%
(iv) Mass of NaCl in the mixture
\[
\text{Mass of NaCl}
=15.0-10.4
=4.6\text{ g dm}^{-3}
\]
Mass of NaCl = 4.6 g dm-3
