# pKa and equivalence point [closed]

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.

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

$$99.9\%$$ 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.