Calculate the volume of ammonium hydroxide needed to get desired pH I have an $29\%$ aqueous solution of ammonium hydroxide. How much do I need to add to $\pu{20 ml}$ of water to get $\mathrm{pH}$ $12.8$. Ammonium hydroxide $K_\mathrm{b}$ is $1.774\cdot 10^{-5}$.


What I tried:

$$K_\mathrm{b} = \frac{[\ce{OH-}][\ce{NH4+}]}{[\ce{NH4OH}]} \approx \frac{[\ce{OH-}]^2}{[\ce{NH4OH}]}$$

Could someone explain the rational for this approximation if it is correct?

Then

$$
\begin{align}
-\log{K_\mathrm{b}} &amp;= -2\log{[\ce{OH-}]} + \log{[\ce{NH4OH}]} \\
\mathrm{p}K_\mathrm{b} &amp;= 2\mathrm{pOH} + \log{[\ce{NH4OH}]} \\
[\ce{NH4OH}] &amp;= 10^{\mathrm{p}K_\mathrm{b} - 2\mathrm{pOH}} \\
&amp;= 10^{\mathrm{p}K_\mathrm{b}-2(\mathrm{p}K_\mathrm{w}-\mathrm{pH})} \\
\mathrm{p}K_\mathrm{b} &amp;= -\log{(1.774\cdot 10^{-5})} = 4.75 \\
\mathrm{p}K_\mathrm{w} &amp;= 14 \\
\mathrm{pH} &amp;= 12.8
\end{align}
$$

Plugging in:

$$[\ce{NH4OH}] = 10^{4.75-2(14-12.8)} = 223.9$$

This is much too high and does not make sense.
Technically, you calculated the concentration correctly (math-wise). However, your initial hypothesis is wrong.

First, you cannot get ammonia hydroxide solution of $\mathrm{pH}$ $12.8$ at the room temperature to work with. At $\pu{20^\circ C}$ concentrated solution would be approx. $35\%$ with $\mathrm{pH} \approx 12.0$. The higher the temperature, the lower the solubility and the $\mathrm{pH}$ will be. Having $29\%$ stock ammonia solution, there is no way to reach $\mathrm{pH}$ $12.8$.

Second, when you work with borderline/concentrated solutions, you are no longer allowed to use concentrations when describing the equilibrium, you need to use activities instead. The formula you derived would give a reasonable result for the moderately concentrated solutions only. You can adapt it to establish the relationship between the mass fraction $\omega$ and $\mathrm{pH}$, knowing the density of the solution $\rho$ ($M$ is molar mass of ammonia)

$$\mathrm{pH} = 14 + 0.5\lg{\frac{\omega\rho K_\mathrm{b}}{M}}\, ,$$

but that's pretty much all to it.