Question

A certain acidic indicator has $pH =5.48. \,75 \%$ of the indicator exists in its ionised from. Calculate $Ka$ of the indicator,

• A. $1 \times 10^{-5}$
• B. $2 \times 10^{-5}$
• C. $1 \times 10^{-3}$
• D. $4 \times 10^{-5}$

Answer 1

Answer: (A)

$pH = pk _{ In }+\log \left(\frac{ In ^{-}}{ H In }\right)$
$5.48= pk _{ In }+\log \left(\frac{75}{25}\right)$
$pk _{ In }=5.48-\log 3=5.48-0.48=5$
$k _{ In }=1 \times 10^{-5}$