1. Indicator equilibrium for phenolphthalein

Phenolphthalein is treated as a weak acid-type indicator:

$\mathrm{HIn \rightleftharpoons H^+ + In^-}$

with

$K_{\text{phen}}=\frac{[\mathrm{H^+}][\mathrm{In^-}]}{[\mathrm{HIn}]}=4\times 10^{-10}$

Hence,

$pK_{\text{phen}}=-\log(4\times 10^{-10})=10-\log 4$

Given $\log 2=0.3$, so

$\log 4=2\log 2=0.6$

Therefore,

$pK_{\text{phen}}=10-0.6=9.4$

For an indicator, visible colour change begins when

$\frac{[\mathrm{In^-}]}{[\mathrm{HIn}]}\approx \frac{1}{10}$

Using Henderson relation:

$pH=pK_{\text{phen}}+\log\frac{[\mathrm{In^-}]}{[\mathrm{HIn}]}$

At the beginning of colour change,

$pH=9.4+\log\left(\frac{1}{10}\right)=9.4-1=8.4$

So statement B is correct.

2. Nature of phenolphthalein

Since the equilibrium is written as acid dissociation,

$\mathrm{HIn \rightleftharpoons H^+ + In^-}$

phenolphthalein behaves as a weak organic acid, not a base.

So statement C is false.

3. Colour in acidic medium

In acidic medium, $[\mathrm{H^+}]$ is high, so equilibrium shifts left and indicator remains mostly in the $\mathrm{HIn}$ form, which is colourless for phenolphthalein.

So statement D is correct.

4. Use in weak acid vs weak base titration

For a weak acid–weak base titration, the pH change near equivalence point is not sharp enough. Hence phenolphthalein is not suitable as indicator.

So statement A is false.

5. Count correct statements

• A: False
• B: True
• C: False
• D: True

Therefore, number of correct statements is

$2$

6. Comparison with stored answer

Stored correct answer = $2$

Our derived answer also = $2$.