Question

An acid indicator $HIn$ has $K_a=3\times {10}^{-5}$, the acid form is red and basic form is blue. Which is correct?

• A. $pH=5$ when indicator is $75$% red
• B. $pH=4.05$ when indicator is $75$% red
• C. $pH=5$ when indicator is $75$% blue
• D. $pH=4.05$ when indicator is $75$% blue

Answer 1

Answer: (B), (C)

$pH=4.05$ when indicator is $75$% red
$pH=5$ when indicator is $75$% blue

$HIn\rightleftharpoons H^{+}+I{n}^{-}$

When indicator is 75% red it means $HIn$=0.75 and $I{n}^{-}$=0.25

$K_a=\frac{\left[H^{+}\right]\left[{In}^{-}\right]}{HIn}$

$3\times {10}^{-5}=\frac{\left[H^{+}\right]\left[0.25\right]}{0.75}$

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

As we know $pH=-log\left(\right[H^{+}\left]\right)=4.05$

When indicator is 75% blue it means $HIn$=0.25 and $I{n}^{-}$=0.75

$K_a=\frac{\left[H^{+}\right]\left[{In}^{-}\right]}{HIn}$

$3\times {10}^{-5}=\frac{\left[H^{+}\right]\left[0.75\right]}{0.25}$

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

As we know $pH=-log\left(\right[H^{+}\left]\right)=5$