An indicator HIn has K_a = 1 × 10^-8. At pH = 6.0, what is the ratio HIn/In^-?

Question

Quizplus · Accepted Answer

The given $K_a = 1 \times 10^{-8}$. The Henderson-Hasselbalch equation is $pH = pK_a + \log\left(\frac{[In^-]}{[HIn]}\right)$. Substituting the given values, $6.0 = 8 + \log\left(\frac{[In^-]}{[HIn]}\right)$, which simplifies to $\log\left(\frac{[In^-]}{[HIn]}\right) = -2$. Therefore, $\frac{[In^-]}{[HIn]} = 10^{-2}$, which means $[HIn]/[In^-] = 100/1$.

An Indicator HIn Has Ka = 1 × 10-8

An Indicator HIn Has K_a = 1 × 10^-8