Question 21 Red cabbage indicator chart Colour red violet purple blue green yellow pH 1 2 3 4 5 6 7 8 9 10 11 12 13 14 (a) State what colour the red cabbage indicator would be in a 0.005 mol L⁻¹ solution of H₂SO₄ - HSC - SSCE Chemistry - Question 21 - 2007 - Paper 1

When 10 mL of 0.005 mol L⁻¹ H₂SO₄ is diluted with water to a total volume of 100 mL, the new concentration can be calculated using the dilution formula:

$C_1V_1 = C_2V_2$
Where:

• $C_1 = 0.005 \text{ mol L}^{-1}$ (initial concentration)
• $V_1 = 10 \text{ mL}$ (initial volume)
• $V_2 = 100 \text{ mL}$ (final volume)
• $C_2$ = ? (final concentration)

Rearranging gives:
$C_2 = \frac{C_1V_1}{V_2} = \frac{0.005 \times 10}{100} = 0.0005 \text{ mol L}^{-1}$

With a concentration of 0.0005 mol L⁻¹, the pH of the solution would be approximately 6, making the colour of the solution green according to the indicator chart.

To neutralise the sulfuric acid, we need to calculate the number of moles of H₂SO₄ in 15 mL. Using the final concentration of H₂SO₄ from the previous part:

For 15 mL (0.015 L) of the diluted H₂SO₄:
$n = C \times V = 0.0005 \text{ mol L}^{-1} \times 0.015 \text{ L} = 7.5 \times 10^{-6} \text{ mol}$

According to the neutralisation reaction: $H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O$

We see that 1 mole of H₂SO₄ reacts with 2 moles of KOH. Therefore, the moles of KOH needed are:
$n_{KOH} = 2n_{H₂SO₄} = 2 \times 7.5 \times 10^{-6} = 1.5 \times 10^{-5} \text{ mol}$

Now, we calculate the volume of KOH solution needed:
$V = \frac{n}{C} = \frac{1.5 \times 10^{-5}}{0.005} = 0.003 ext{ L} = 3 ext{ mL}$

Thus, 3 mL of 0.005 mol L⁻¹ KOH is required to neutralise 15 mL of the diluted H₂SO₄.