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An acid – base indicator has K_a = 3 × 10^–5. The acid form of indicator is red and the basic form is blue. The [H⁺] required to change the indicator from 75% red to 75% blue is

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8 x 10^-5

9 x10^-5

10^-5

3x 10^-4

answer is A.

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Detailed Solution

	HIn	$⇌$ H⁺	+In^-
	red		blue
t₁	75		$25 \to P^{H_{1}} = 5 - \log 3 + \log \frac{1}{3}$
t₂	25		$75 \to P^{H_{2}} = 5 - \log 3 + \log 3$

$P^{H_{1}} = 5 - 2 \log 3 = 5 - \log 3^{2}$

$[H^{+}] = 9 \times 10^{- 5}$

$P^{H_{2}} = 5$

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

$Δ [H^{+}] = 10^{- 5} (9 - 1) = 8 \times 10^{- 5}$

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