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

To complete Figure 11, plot each point from Table 6 on the graph with the x-axis representing the volume of dilute sulfuric acid added (cm³) and the y-axis representing the temperature (°C). After plotting all six points, draw lines of best fit for the data, ensuring that the lines extend until they cross at the point where the volumetric reaction is complete.

From Figure 11, find the intersection of the two lines of best fit. The volume of dilute sulfuric acid needed to react completely with 25.0 cm³ of potassium hydroxide solution is 11 cm³.

The overall temperature change can be determined by finding the temperature of the potassium hydroxide solution before adding the acid and the final temperature after the complete addition of sulfuric acid. From the values provided, the temperature change is calculated as:

Overall temperature change = 27.0 °C - 18.9 °C = 8.1 °C.

To calculate the concentration of the potassium hydroxide solution:

1. From the balanced equation, we know:

   $2 \text{ KOH} + \text{ H}_2\text{SO}_4 \rightarrow \text{ K}_2\text{SO}_4 + 2 \text{ H}_2\text{O}$

   For 15.5 cm³ of 0.500 mol/dm³ sulfuric acid:

   • Moles of H2SO4 = 0.5 * (15.5 / 1000) = 0.00775 moles
   • Moles of KOH = 2 * moles of H2SO4 = 2 * 0.00775 = 0.0155 moles

2. Concentration (C) of KOH:

   $C = \frac{moles\ of\ KOH}{volume\ in\ dm³} = \frac{0.0155}{25.0 \times 10^{-3}} = 0.62 \text{ mol/dm}^3$

3. To convert to g/dm³:

   • Molar mass of KOH = 56 g/mol, thus:

   $concentration\ in\ g/dm^3 = moles \times molar\ mass = 0.62 \times 56 = 34.72\ g/dm^3$