All your burette readings (initial and final), as well as the size of your pipette, must be recorded but no account of the experimental procedure is required. All calculations must be done in your answer booklet.
A is a solution containing 5.00 g of HNO\(_3\) in 500 cm\(^3) of solution. B is a solution of NaOH of unknown concentration.
(a) Put A into the burette and titrate it with 20.0 cm\(^3\) or 25.0 cm\(^3\) portions of B using methyl orange as an indicator. Repeat the titration to obtain concordant titre values. Tabulate your results and calculate the average volume of acid used. Equation of the reaction is HNO\(_{3(aq)}\) + NaOH\(_{(aq)}\) \(\to\) NaNO\(_{3(aq)}\) + H\(_2\)O\(_{(l)}\)
(b) From your results and the information provided. calculate the: (i) concentration ot A In mol dm\(^{-3}\)
(ii) concentration of B in mol dm\(^{-3}\).
(iii) concentration of B in gdm\(^{-3}\)
(iv) mass of NaNO\(_3\) formed. If 250 cm\(^3\) of NaOH were neutralised. [Molar mass of NaOH = 40g mol\(^{-1}\), NaNO\(_3\) = 85 gmol\(^{-1}\). Credit will be given for strict adherence to the instructions. for observations precisely recorded and for accurate inferences. All tests, observations and inferences must be clearly entered in this booklet, in ink, at the time they are made.
A is a solution containing 5.00 g of HNO\(_3\) in 500 cm\(^3) of solution. B is a solution of NaOH of unknown concentration.
(a) Put A into the burette and titrate it with 20.0 cm\(^3\) or 25.0 cm\(^3\) portions of B using methyl orange as an indicator. Repeat the titration to obtain concordant titre values. Tabulate your results and calculate the average volume of acid used. Equation of the reaction is HNO\(_{3(aq)}\) + NaOH\(_{(aq)}\) \(\to\) NaNO\(_{3(aq)}\) + H\(_2\)O\(_{(l)}\)
(b) From your results and the information provided. calculate the: (i) concentration ot A In mol dm\(^{-3}\)
(ii) concentration of B in mol dm\(^{-3}\).
(iii) concentration of B in gdm\(^{-3}\)
(iv) mass of NaNO\(_3\) formed. If 250 cm\(^3\) of NaOH were neutralised. [Molar mass of NaOH = 40g mol\(^{-1}\), NaNO\(_3\) = 85 gmol\(^{-1}\). Credit will be given for strict adherence to the instructions. for observations precisely recorded and for accurate inferences. All tests, observations and inferences must be clearly entered in this booklet, in ink, at the time they are made.
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Correct Answer: Option
Explanation:
(a)
The first titration as a trial
Average volume of acid used
= \(\frac{16.10 + 16.10cm^3}{2}\)
= \(\frac{32.20cm^3}{2}\)
= 16.10cm\(^3\)
(b)(i) Concentration of A in mol/dm\(^3\)
500cm\(^3\) of solution contains 5.00g of HNO\(_3\)
100cm\(^3\) of solution will contain
\(\frac{5.00}{500} \times \frac{100}{1}\) = 10.00 gdm\(^{-3}\)
Molar mass of HNO\(_3\) = (1 x 1) + (14 x 1) + (16 x 3)
= 1 + 14 + 48 = 63gmol\(^{-1}\)
Using the relation:
Concentration g/dm\(^3\)
= molar conc.moldm\(^{-3}\) x molar mass
10.00 = molar conc. moldm\(^{-3}\) x 63
Concentration of A = \(\frac{10.00}{63}\)
= 0.15873015873 moldm\(^{-3}\)
= 0.158 moldm\(^{-3}\)
(ii) Concentration of B in moldm\(^{-3}\)
Using the formula
\(\frac{C_AV_A}{C_BV_B} = \frac{N_A}{N_B}\)
\(\frac{0.159 \times 16.10}{C_B \times 25.0} = \frac{1}{1}\)
\(C_B = \frac{0.159 \times 16.10 \times 1}{25.0 \times 1}\)
= 0.0102396 moldm\(^{-3}\)
\(C_B\) = 0.102 moldm\(^{-3}\)
(iii) Concentration of B in gdm\(^{-3}\)
concentration g/dm\(^{-3}\) = molar concentration of moldm\(^{-3}\) x molar mass
Molar mass NaOH = 40g/mol
Concentration of B = 0.102 x 40
= 4.08g/dm\(^3\)
(iv) Mass of NaOH
No. of moles of NaOH = \(\frac{\text{molar conc.moldm}^{-3} \times volume}{100}\)
= \(\frac{0.102 \times 250}{1000}\) = 0.0255 mole
from the equation
1 mole of NaOH = 0.055mole of NaNo\(_3\)
0.0255 mole of NaOH = 0.0255 mole of NaNO\(_3\)
molar mass of NaNO\(_3\) = 85 gmol\(^{-1}\)
mass of NaNO\(_3\) = 0.0255 x 85
= 2.1675g
(a)
| Burette reading/Titration | 1st Titration | 2nd Titration | 3rd Titration |
| Final burette reading (cm\(^3\)) | 16.50 | 16.10 | 21.10 |
| Initial burette reading (cm\(^3\)) | 0.00 | 0.00 | 5.00 |
| Average volume of acid used | 16.50 | 16.10 | 16.10 |
The first titration as a trial
Average volume of acid used
= \(\frac{16.10 + 16.10cm^3}{2}\)
= \(\frac{32.20cm^3}{2}\)
= 16.10cm\(^3\)
(b)(i) Concentration of A in mol/dm\(^3\)
500cm\(^3\) of solution contains 5.00g of HNO\(_3\)
100cm\(^3\) of solution will contain
\(\frac{5.00}{500} \times \frac{100}{1}\) = 10.00 gdm\(^{-3}\)
Molar mass of HNO\(_3\) = (1 x 1) + (14 x 1) + (16 x 3)
= 1 + 14 + 48 = 63gmol\(^{-1}\)
Using the relation:
Concentration g/dm\(^3\)
= molar conc.moldm\(^{-3}\) x molar mass
10.00 = molar conc. moldm\(^{-3}\) x 63
Concentration of A = \(\frac{10.00}{63}\)
= 0.15873015873 moldm\(^{-3}\)
= 0.158 moldm\(^{-3}\)
(ii) Concentration of B in moldm\(^{-3}\)
Using the formula
\(\frac{C_AV_A}{C_BV_B} = \frac{N_A}{N_B}\)
\(\frac{0.159 \times 16.10}{C_B \times 25.0} = \frac{1}{1}\)
\(C_B = \frac{0.159 \times 16.10 \times 1}{25.0 \times 1}\)
= 0.0102396 moldm\(^{-3}\)
\(C_B\) = 0.102 moldm\(^{-3}\)
(iii) Concentration of B in gdm\(^{-3}\)
concentration g/dm\(^{-3}\) = molar concentration of moldm\(^{-3}\) x molar mass
Molar mass NaOH = 40g/mol
Concentration of B = 0.102 x 40
= 4.08g/dm\(^3\)
(iv) Mass of NaOH
No. of moles of NaOH = \(\frac{\text{molar conc.moldm}^{-3} \times volume}{100}\)
= \(\frac{0.102 \times 250}{1000}\) = 0.0255 mole
from the equation
1 mole of NaOH = 0.055mole of NaNo\(_3\)
0.0255 mole of NaOH = 0.0255 mole of NaNO\(_3\)
molar mass of NaNO\(_3\) = 85 gmol\(^{-1}\)
mass of NaNO\(_3\) = 0.0255 x 85
= 2.1675g