Water and octan-1-ol do not mix - HSC - SSCE Chemistry - Question 39 - 2024 - Paper 1

To calculate the equilibrium constant K_eq, we need the expression:

$K_{eq} = \frac{[H^+] [BrCH_2COOH]}{[BrCH_2COOH(aq)]}$

Substituting the known values into the formula:

$K_{eq} = \frac{[9.18 \times 10^{-3}] \times [BrCH_2COOH]}{[BrCH_2COOH(aq)]}$

From the initial concentration, the change can be calculated:

• Initial concentration of bromoacetic acid: 0.1000 mol L⁻¹
• Since [H⁺] = 9.18 x 10⁻³ mol L⁻¹ at equilibrium, the concentration of bromoacetic acid that dissociated is approximately 0.0000.
• Therefore, the concentration of bromoacetic acid at equilibrium:

$[BrCH_2COOH(aq)] = 0.1000 - 9.18 \times 10^{-3} = 0.0654 \text{ mol L}^{-1}$

Now substituting these values back into the K_eq expression gives:

$K_{eq} = \frac{[9.18 \times 10^{-3}] \times [0.0654]}{0.1000} = 0.390$