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

• A. $3 \text{.} 5 \times 1 0^{- 4}$
• B. $3 \text{.} 5 \times 1 0^{- 5}$
• C. $3 \text{.} 5 \times 1 0^{- 2}$
• D. $3 \text{.} 5 \times 1 0^{- 3}$

Answer 1

Answer: (A)

For weak organic acid indicators

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

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

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

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

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

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