What fraction of an indicator (Hln) is in basic form at a pH of 6 if the p K a of the indicator is 5 ?

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What fraction of an indicator (Hln) is in basic form at a pH of 6 if the

p K_{a}

of the indicator is 5 ?

ChemistryIonic EquilibriumNEET

Options:

A $\frac{1}{2}$
B $\frac{1}{11}$
C $\frac{10}{11}$
D $\frac{1}{10}$

Solution:

1338 Upvotes Verified Answer

The correct answer is:

\frac{10}{11}

p H = p K_{a} + l o g \frac{(I n^{-})}{(H I n)}

6 = 5 + l o g \frac{(I n^{-})}{(H I n)}

\frac{(I n^{-})}{(H I n)} = 10

\frac{(I n^{-})}{(I n^{-}) + (H I n)} = \frac{10}{11}

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