For an acid indicator having the formula $H I n$ . Choose the correct options when the ratio $\frac{[I n^{-}]}{[H I n]} = 10$

p H = p K_{I n} + 1

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% [I n^{-}] = 90.9 %

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p H = p K_{I n} - 1

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% [H I n] = 90.9 %

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Solution

The correct options are
A $p H = p K_{I n} + 1$
B $% [I n^{-}] = 90.9 %$
$H I n (a q)_{I n} H^{+} (a q) + I n^{-} (a q)$

$K_{I n} = \frac{[H^{+}] [I n^{-}]}{[H I n]}$
$[H^{+}] = \frac{K_{I n} [H I n]}{[I n^{-}]}$
$p H = p K_{I n} + l o g \frac{[I n^{-}]}{[H I n]}$
$p H = p K_{I n} + l o g \frac{[Ionised form]}{[Unionised form]}$
When,

$\frac{[I n^{-}]}{[H I n]} = 10 = \frac{[I o n i s e d]}{[U n i o n i s e d]}$

$p H = p K_{I n} + l o g (10) p H = p K_{I n} + 1$
Option A is correct.
Also,

$[I n^{-}] = 10 [H I n]$
$% [I n^{-}] = \frac{[I n^{-}]}{[H I n] + [I n^{-}]} \times 100$
$% [I n^{-}] = \frac{10}{1 + 10} \times 100 = 90.9 %$
$% H I n = (100 - 90.9) % = 9.1 %$
Option B is correct.

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