Question

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: $log7=0.845$ and Antilog $(-3.455)=3.5\times {\left(10\right)}^{-4}$ ]

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

Answer 1

Answer: (A)

For weak organic acid indicators

$\text{pH}={\text{pK}}_{\text{In}}+\text{log}\frac{\left[{\text{In}}^{-}\right]}{\left[\text{HIn}\right]}$

$\text{4}\text{.3}={\text{pK}}_{\text{In}}+\text{log 7}$

${\text{pK}}_{\text{In}}=4.3-0.845=3.455$

${\text{pK}}_{\text{In}}=-{\text{log}}_{\text{10}}{\text{K}}_{\text{In}}$

${\mathrm{K}}_{\mathrm{In}}=\mathrm{Antilog}\left(-{\mathrm{pK}}_{\mathrm{In}}\right)$
${\mathrm{K}}_{\mathrm{In}}=$ Antilog $(-3.455)$

$=3.5\times 10^{-4}$