How can I choose an indicator for determining the concentration of a base using its pKb values?
Since: 
\begin{align}
K_\mathrm{w} &amp;=K_\mathrm{a} \cdot K_\mathrm{b}\\
K_\mathrm{a} &amp;=\frac{K_\mathrm{w}}{K_\mathrm{b}}\\
\log{K_\mathrm{a}} &amp;=\log K_\mathrm{w}-\log K_\mathrm{b}\\
K_\mathrm{w} &amp;=1 \times 10^{-14}\\
\log{K_\mathrm{a}} &amp;= -14-\log K_\mathrm{b}\\
-\log{K_\mathrm{a}} &amp;=14+\log K_\mathrm{b}\\
-\log{K_\mathrm{a}} &amp;=\mathrm{p}K_\mathrm{a}\\
-\log{K_\mathrm{b}} &amp;=\mathrm{p}K_\mathrm{b}\\
\mathrm{p}K_\mathrm{a} &amp;=14-\mathrm{p}K_\mathrm{b}
\end{align}

Using the above formula, the three $\mathrm{p}K_\mathrm{a}$ for the conjugate acid of the conjugate base are: $13.5~(\mathrm{p}K_\mathrm{a_1})$, $10.3~(\mathrm{p}K_\mathrm{a_2})$, and $6.3~(\mathrm{p}K_\mathrm{a_3})$, where $\mathrm{p}K_\mathrm{a_1}=14-\mathrm{p}K_\mathrm{b_1}$, etc.

An important equation to note is that the $\mathrm{pH}$ at the equivalence point is given by the equation: $$\mathrm{pH_\mathrm{eq}}=\frac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{2}$$

This means that two of the equivalence points exist at $$\mathrm{pH_\mathrm{eq_1}}=\frac{13.5+10.3}{2}=11.9$$ and $$\mathrm{pH_\mathrm{eq_2}}=\frac{10.3+6.3}{2}=8.3$$

The $\mathrm{pH}$ range for methyl orange is $3.0~-~4.4$ and the $\mathrm{pH}$ range for phenolphthalein is $8.2~-~10.0$. Phenolphthalein would be useful for detecting the second equivalence point $(\mathrm{pH_\mathrm{eq_2}})$, since this shift occurs within phenolphthalein's range.

The final equivalence point cannot be calculated from the information given, but given that methyl orange's color transition occurs in an acidic medium, and also the relatively basic $\mathrm{p}K_\mathrm{a_1}$, methyl orange can be eliminated as a possible indicator.