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A student investigated the temperature change in the reaction between dilute sulfuric acid and potassium hydroxide solution - AQA - GCSE Chemistry - Question 9 - 2019 - Paper 1

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A student investigated the temperature change in the reaction between dilute sulfuric acid and potassium hydroxide solution. This is the method used. 1. Measure 25... show full transcript

Worked Solution & Example Answer:A student investigated the temperature change in the reaction between dilute sulfuric acid and potassium hydroxide solution - AQA - GCSE Chemistry - Question 9 - 2019 - Paper 1

Step 1

Suggest why the student used a polystyrene cup rather than a glass beaker for the reaction.

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Answer

Polystyrene is a better (thermal) insulator than glass. This property reduces energy exchange with the surroundings, minimizing heat loss during the reaction.

Step 2

Complete Figure 11: plot the data from Table 6; draw a line of best fit through these points; extend the lines of best fit until they cross.

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Answer

Plot the temperature changes against the volume of dilute sulfuric acid added from Table 6 on a graph. Draw a line of best fit through the plotted points, ensuring all six points from Table 6 are plotted accurately. Extend both lines of best fit until they intersect, providing a clear point indicating the volume for complete reaction.

Step 3

Determine the volume of dilute sulfuric acid needed to react completely with 25.0 cm³ of the potassium hydroxide solution.

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Answer

From the graph in Figure 11, the volume of dilute sulfuric acid needed to completely react with 25.0 cm³ of potassium hydroxide solution is found to be approximately 11 cm³.

Step 4

Determine the overall temperature change when the reaction is complete.

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Answer

To find the overall temperature change, subtract the initial temperature (18.9 °C) from the final temperature (27.1 °C) observed when the reaction is complete. The overall temperature change is 27.1 °C - 18.9 °C = 8.2 °C.

Step 5

Calculate the concentration of the potassium hydroxide solution in mol/dm³ and in g/dm³.

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Answer

First, calculate the moles of moles H2SO4=0.500 mol/dm3×15.5 cm31000=0.00775 molesmoles \ H_2SO_4 = 0.500 \ mol/dm^3 \times \frac{15.5 \ cm^3}{1000} = 0.00775 \ moles.

Since 2 moles of KOH react with 1 mole of H₂SO₄, the moles of KOH are: moles KOH=2×0.00775=0.0155 molesmoles \ KOH = 2 \times 0.00775 = 0.0155 \ moles.

To find the concentration of KOH: conc KOH=moles KOH×100025.0 cm3=0.0155 times100025.0=0.62 mol/dm3conc \ KOH = \frac{moles \ KOH \times 1000}{25.0 \ cm^3} = \frac{0.0155 \ times 1000}{25.0} = 0.62 \ mol/dm^3.

Then, to find the mass concentration in g/dm³, use the molar mass of KOH (56 g/mol): conc=56×0.62=34.72 g/dm3conc = 56 \times 0.62 = 34.72 \ g/dm^3.

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