1
$\begingroup$

I was doing a titration and using $\ce{NaOH}$ as the titrant. The analyte was a weak acid with a $\mathrm{p}K_\mathrm{a}$ of 4.90. Today I want to use another acid with a $\mathrm{p}K_\mathrm{a}$ of 4.50. Do I need to change the indicator from the last experiment? How close do the $K_\mathrm{a}$ or $\mathrm{p}K_\mathrm{a}$ values have to be in order for me to use the same indicator and still obtain accurate results.

$\endgroup$

closed as off-topic by airhuff, Buttonwood, Todd Minehardt, Mithoron, Nuclear Chemist May 8 at 17:45

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ It's not a matter of their closeness; it is a matter of indicator's properties. $\endgroup$ – Ivan Neretin May 7 at 20:30
1
$\begingroup$

$99.9\%$ of the acid $\ce{HA}$ is titrated at $\mathrm{pH}=7.9$ resp.$ 7.5$.

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a}+\log \frac{[\ce{A-}]}{[\ce{HA}]}=\mathrm{p}K_\mathrm{a}+3$$

Considering the final concentration 0.05M,
titrating 0.1M acid by 0.1M hydroxide,
100% of the acid is titrated at:

$$\begin{align} \mathrm{pH}&=7+\frac12 \cdot ( \mathrm{p}K_\mathrm{a} + \log c_\mathrm{acid, total}) \\ &=9.45(9.25) + \frac12 \cdot (-1.3) \\ &=8.8(8.6)\\ \end{align}$$

Titration with $100.1\%$ of used hydroxide we can consider
as $0.05 \cdot 0.001=5\cdot 10^{-5}\ \mathrm{M} \ \ce{NaOH}$
with $\mathrm{pH}=-\log ( 5\cdot 10^{-5})=9.7$

The indicator phenolphthalein with range $8.2-10.0$ works well in both cases.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.