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Thermometric Titration – Answers
Question: A student investigated the rise in temperature when sulfuric acid was added to a solution containing 1.00 mol/dm3 sodium hydroxide, using the apparatus shown below:
3 3 20.0 cm of 1.00 mol/dm sodium hydroxide was poured into a beaker. The initial temperature (Ti) of both this solution and the sulfuric acid was 25.0C.
Next, 5.0 cm3 of sulfuric acid was added to the aqueous sodium hydroxide from the burette. The reaction mixture was stirred gently and the maximum temperature (Tm) was taken. Following successive additions of 5.0 cm3 volumes of sulfuric acid from the burette, further temperature readings (Tm) were taken.
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The diagrams below show parts of the thermometer stem giving the temperature after the addition of 5.0, 15.0 and 25.0 cm3 of sulfuric acid.
a) i) Use the diagrams to complete the following table of results. [1]
3 ii) Calculate the change in temperature (Tm – Ti) for each 5.0 cm volume of sulfuric acid added to the aqueous sodium hydroxide. Complete this on the table of results. [1]
Volume of Sulfuric Maximum Temperature Change in Temperature 3 Acid / cm Tm / C Tm – Ti / C
5.0 27.0 2.0
10.0 29.0 4.0
15.0 31.0 6.0
20.0 33.0 8.0
25.0 32.0 7.0
30.0 29.0 4.0
35.0 26.0 1.0
2 b) Plot the change in temperature, Tm – Ti against volume of sulfuric acid on the grid below. Connect the points with two intersecting straight lines.
[3]
Use the graph to answer the following questions. c) i) State the change in temperature at the intersection of the two lines. 8.8 C (9.0 C to the nearest whole degree Celsius) ii) What volume of sulfuric acid produced this temperature? 22.0 cm3 [2] d) 20.0 cm3 of 1.00 mol/dm3 sodium hydroxide was used in the experiment. i) Write an equation for the reaction between sodium hydroxide and sulfuric acid.
2NaOH (aq) + H2SO4 (aq) Na2SO4 (aq) + 2H2O (l) [1]
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ii) Using your answer to c) ii), calculate the concentration of the sulfuric acid. mol = c v 10–3 mol of NaOH = 1.00 20.0 10–3 mol of NaOH = 0.0200 mol
from the balanced chemical equation, 2 mol of NaOH reacts with 1 mol H2SO4
1 0.0200 mol NaOH reacts with /2 0.0200 = 0.0100 mol H2SO4 c = (mol v) 1000
3 concentration of H2SO4 = (0.0100 22.0) 1000 = 0.455 mol/dm [2] e) Use the formula given below to calculate the enthalpy change of this reaction to three significant figures. H = m c T H = enthalpy change / J m = mass of solution / g c = specific heat capacity of water = 4.18 J/g/C T change in temperature / C Note: Assume the density of the solution = 1.00 g/cm3 T = 8.80 C
3 volume of solution at end-point = volume NaOH + volume H2SO4 = 20.0 + 22.0 = 42.0 cm mass of solution at end-point = density volume = 1.00 g/cm3 42.0 cm3 = 42.0 g H = 42.0 4.18 8.80 = +1544.928 J H = +1540 J to 3 s.f. Note: This is energy gained by the surroundings causing an increase in temperature, therefore the actual enthalpy change for the reaction is –1540 J. [3] f) After the highest temperature was reached, explain why the temperature of the solution decreased as more sulfuric acid was added. Sulfuric acid at room temperature (25.0 C) is added to the solution in excess, therefore cooling the solution down. The solution loses heat to the surroundings. [2] g) Identify a possible source of error for this experiment and clearly state how the error affects the results of the experiment. Heat loss to the surroundings will cause the value of T to be smaller than it should be. This will cause the calculated value of H to smaller than it should be. [2] [Total: 17] 4