Fig. 7.1 Fig. 7.1 shows the apparatus used to titrate a solution of sodium carbonate against dilute nitric acid. (a) State how the burette reading should be...
Question 1 Report
Fig. 7.1
Fig. 7.1 shows the apparatus used to titrate a solution of sodium carbonate against dilute nitric acid.
(a) State how the burette reading should be taken so that a parallax error is avoided. [1]
(b) State why a conical flask is used to hold the sodium carbonate solution rather than an open beaker. [1]
(c) Methyl orange is used as the indicator. State the colour change seen in the flask as the acid is added to the sodium carbonate solution up to the end-point. [2]
(d) A pipette delivers 20.0 cm3 of sodium carbonate solution, concentration 0.0500 mol/dm3, into the flask. The average titre of nitric acid is 16.00 cm3. The equation is:
Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2
(i) Calculate the amount, in mol, of sodium carbonate used. [1]
(ii) Calculate the amount, in mol, of nitric acid used. [1]
(iii) Calculate the concentration of the nitric acid in mol/dm3. [2]
(e) Calculate the concentration of the nitric acid in g/dm3. [2]
(f) Describe a test, and its positive result, to show that the gas given off is carbon dioxide. [2]
(g) Name the salt formed in this titration. [1]
Answer Details
Sodium carbonate is titrated against nitric acid. The mole ratio from the equation is 1:2, and the question also tests practical technique, indicator colour, converting mol/dm3 to g/dm3, and the test for carbon dioxide.
(a) Avoiding parallax error [1]. Read the scale with your eye level with the bottom of the meniscus [1], so the line of sight is horizontal.
(b) Why a conical flask, not a beaker [1]. A conical flask can be swirled vigorously to mix the reagents without any solution splashing out, so no liquid is lost [1].
(c) Methyl orange colour change [2]. As acid is added to the alkaline carbonate the indicator changes from yellow [1] to orange/red at the end-point [1].
(d)(i) Moles of sodium carbonate [1]. \( n(\text{Na}_2\text{CO}_3) = c \times V = 0.0500 \times \dfrac{20.0}{1000} = 0.00100\ \text{mol} \) [1].
(d)(ii) Moles of nitric acid [1]. The equation shows 1 Na2CO3 reacts with 2 HNO3, so \( n(\text{HNO}_3) = 2 \times 0.00100 = 0.00200\ \text{mol} \) [1].
(d)(iii) Concentration of nitric acid [2]. This is in the 16.00 cm3 titre, so \( c = \dfrac{n}{V} = \dfrac{0.00200}{16.00/1000} = 0.125\ \text{mol/dm}^3 \) [2].
(e) Concentration in g/dm3 [2]. First the molar mass: \( M_r(\text{HNO}_3) = 1 + 14 + (3 \times 16) = 63 \) [1]. Then \( 0.125\ \text{mol/dm}^3 \times 63\ \text{g/mol} = 7.875\ \text{g/dm}^3 \) [1].
(f) Test for carbon dioxide [2]. Bubble the gas through limewater (calcium hydroxide solution) [1]; a positive result is the limewater turning milky/cloudy as a white precipitate of calcium carbonate forms [1].
(g) Salt formed [1]. Sodium nitrate (NaNO3) [1].
Exam tip: to change mol/dm3 into g/dm3, multiply by the molar mass; to go back, divide.
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