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 require...
Question 1 Report
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 \( \mathrm{HNO}_3 \) 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 \( \mathrm{HNO}_{3(aq)} + \mathrm{NaOH}_{(aq)} \to \mathrm{NaNO}_{3(aq)} + \mathrm{H}_2\mathrm{O}_{(l)} \)
(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.
Answer Details
(a) Titration results (pipette volume = 25.0 cm3 of B; indicator methyl orange)
| Trial | 1st | 2nd | |
|---|---|---|---|
| Final reading (cm3) | 16.50 | 16.10 | 16.10 |
| Initial reading (cm3) | 0.00 | 0.00 | 0.00 |
| Volume of A used (cm3) | 16.50 | 16.10 | 16.10 |
Average volume of acid A used = (16.10 + 16.10) / 2 = 16.10 cm3.
(b)(i) Concentration of A in mol dm-3
A contains 5.00 g in 500 cm3, that is 10.0 g dm-3. Molar mass of HNO3 = 1 + 14 + 48 = 63 g mol-1.
[A] = 10.0 / 63 = 0.159 mol dm-3.
(b)(ii) Concentration of B in mol dm-3
For HNO3 + NaOH gives NaNO3 + H2O the mole ratio is 1 : 1.
n(A) = 0.159 × 16.10 / 1000 = 2.56 × 10-3 mol, so n(NaOH) in 25.0 cm3 = 2.56 × 10-3 mol.
[B] = 2.56 × 10-3 / (25.0 / 1000) = 0.102 mol dm-3.
(b)(iii) Concentration of B in g dm-3
= 0.102 × 40 = 4.08 g dm-3.
(b)(iv) Mass of NaNO3 formed from 250 cm3 of NaOH
n(NaOH) = 0.102 × 250 / 1000 = 0.0255 mol. Mole ratio NaOH : NaNO3 = 1 : 1, so n(NaNO3) = 0.0255 mol.
Mass of NaNO3 = 0.0255 × 85 = 2.17 g.
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