Bromophenol blue is a weak acid (represented as BH) that acts as an acid-base indicator - VCE - SSCE Chemistry - Question 2 - 2007 - Paper 1

At the point where [BH] = [B^-], we have:

$K_a = \frac{[B^-][H^+]}{[BH]} = \frac{[BH][H^+]}{[BH]} = [H^+]$

Since [BH] = [B^-], substituting in yields:

$K_a = [H^+]$

Thus, we can find [H^+] using the value of K_a:

$[H^+] = 6.3 \times 10^{-5}$

To find the pH, we calculate:

$pH = -\log([H^+]) = -\log(6.3 \times 10^{-5}) \approx 4.20$

At pH 7,

$[H^+] = 10^{-7} \: M$

Using the K_a expression:

$6.3 \times 10^{-5} = \frac{[B^-](10^{-7})}{[BH]}$

Letting [BH] = x and [B^-] = 6.3 × 10^{-5} / (10^{-7}), we find:

$[B^-] = 630 \: x$

Thus:

$\frac{[B^-]}{[BH]} = \frac{630 \: x}{x} = 630$

At pH 7, the ratio of [B^-]/[BH] is significantly greater than 1 (630:1), which indicates that the solution will predominantly contain the conjugate base, B^-. Therefore, the solution will appear blue. The justification is rooted in the color transition of bromophenol blue, where the presence of more conjugate base will shift the colour towards blue.