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

The correct option is C $1.20$

$pH=pKa+\log \frac{\left[I{n}^{-}\right]}{\left[HIn\right]}$

Initial concentration of $\left[HIn\right]=80$%

Initial concentration of $\left[I{n}^{-}\right]=20$%

Final concengtration of $\left[HIn\right]=20$%

Final concentration of $\left[I{n}^{-}\right]=80$%

Thus,

$p{H}_1=-\log (1\times {10}^{-6})+\log 80/20$

$p{H}_1=6+0.602$

$=6.602$

$p{H}_2=-\log (1\times {10}^{-6})+\log \frac{20}{80}$

$=+6-0.602$

$=5.398$

$pH$ change=$6.602-5.398$

=$1.204$