For the indicator, HIn; the ratio $\frac{\left[{\text{In}}^{-}\right]}{\left[\text{HIn}\right]}$  is $7.0$ at pH of $4.3$. ${\text{K}}_{\text{In}}$ for the indicator is

[ Given: $\log 7=0.845$ and Antilog $\left(-3.455\right)=3.5\times {10}^{-4}$ ]

• Correct answer
  A

  $3\text{.}5\times 10^{-4}$
• B

  $3\text{.}5\times 10^{-5}$
• C

  $3\text{.}5\times 10^{-2}$
• D

  $3\text{.}5\times 10^{-3}$

Hence, the correct answer is option: 'A'.