1. Tardigrade
  2. Question
  3. Chemistry
  4. The ionization constant of an acid-base indicator (a weak acid) is 1.0 × 10-6 . The ionized form of the indicator is red whereas the unionized form is blue. The pH change required to alter the colour of the indicator from 80 % blue to 80 % red is

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Q. The ionization constant of an acid-base indicator (a weak acid) is $1.0 \times 10^{-6} .$ The ionized form of the indicator is red whereas the unionized form is blue. The pH change required to alter the colour of the indicator from $80 \%$ blue to $80 \%$ red is

Equilibrium

Solution:

$HIn \rightleftharpoons H ^{+}+ In ^{-} ; pH = p K_{ In }+\log \frac{\left[ In ^{-}\right]}{[ HIn ]}$

$( pH )_{1}= p K_{ Ln }+\log \frac{20}{80}= p K_{ In }-2 \log 2$

$( pH )_{2}= p K_{ In }+\log \frac{80}{20}= p K_{ In }+2 \log 2$

Hence, $( pH )_{2}-( pH )_{1}= p K_{ In }+2 \log 2-\left( p K_{ In }-2 \log 2\right)$

$=4 \log 2=1.20$