Question

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

• A. 2.00
• B. 1.40
• C. 1.20
• D. 0.80

Answer 1

Answer: (C)

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

$(pH{)}_1=p{K}_{Ln}+\log \frac{20}{80}=p{K}_{In}-2\log 2$

$(pH{)}_2=p{K}_{In}+\log \frac{80}{20}=p{K}_{In}+2\log 2$

Hence, $(pH{)}_2-(pH{)}_1=p{K}_{In}+2\log 2-\left(p{K}_{In}-2\log 2\right)$

$=4\log 2=1.20$