K_w at 25^oC = 1 × 10^–14

At 25ºC

K_w = [H⁺] [OH^–] = 10^–14

At 100°C (given)

K_w = [H⁺] [OH^–] = 55 × 10^–14

for a neutral solution

[H⁺] = [OH^–]

[H⁺]² = 55 × 10^–14

or [H⁺] = (55 × 10–14)^1/2

pH = – log [H⁺]

On taking log on both side

– log [H⁺] = –log (55 × 10^–14)^1/2

$$ \Rightarrow $$ pH = $${1 \over 2}$$ - log55 - 14log10

$$ \Rightarrow $$ pH = 6.13

$$\alpha = \sqrt {{{{K_a}} \over c}} $$

$$ \Rightarrow $$ 0.037 = $$\sqrt {{{{K_a}} \over {0.10}}} $$

$$ \Rightarrow $$ K_a = (0.037)² $$ \times $$ 0.10

= 1.37 ×10^–4

$$ \simeq $$ 1.4 $$ \times $$ 10^$$-$$4

BF₃ acts as Lewis acid. It is electron pair acceptor.

KOH is the base thus, it gives OH^– ions thus it cannot remove OH^– ions from reaction mixture but it adds on the concentration of OH^– ions.

So, an acid must be added but if a strong acid is added to the reaction mixture then in acidic condition the MnO₄ ^– formed reduces to give Mn²⁺ thus, HCl which is a strong acid and SO₂ which on treating with water forms a strong H₂SO₄ cannot be used for this purpose.

Thus, CO₂ which forms H₂CO₃ a weak acid reacts to remove OH^– but not that much acidic that MnO₄ ^– undergo reduction. Thus, CO₂ is used for this reaction for completion.