Ionic EquilibriumHard
Question
An acid–base indicator has Ka = 3.0 × 10−5. The acid form of the indicator is red and the basic form is blue. The [H+] required to change the indicator from 75% blue to 75% red is
Options
A.8 × 10−5 M
B.9 × 10−5 M
C.1 × 10−5 M
D.3.33 × 10−5 M
Solution
$\underset{\text{Red}}{HA} \rightleftharpoons H^{+} + \underset{\text{Blue}}{A^{-}}$
$\left\lbrack H^{+} \right\rbrack = \frac{K_{a}.\lbrack HA\rbrack}{\left\lbrack A^{-} \right\rbrack} $$$\therefore\left\lbrack H^{+} \right\rbrack_{\text{Required}} = \left\lbrack H^{+} \right\rbrack_{2} - \left\lbrack H^{+} \right\rbrack_{1} = K_{a}\left( \frac{75}{25} - \frac{25}{75} \right) = 8 \times 10^{- 5}\text{ M}$$
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