Calculate the volume of ammonium hydroxide needed to get desired pH


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> 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 $1.774\cdot 10^{-5}$.
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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


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


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.


 2

Firstly, "$\ce{NH4OH}$" doesn't really exist. Ammonia solutions generate hydroxide ions via:


$$\ce{NH3(aq) + H2O(l) -> NH4+(aq) + OH-(aq)}$$


The equilibrium constant $K\_b$ is given by:


$$K\_b=\frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}$$


Now let the *nominal* concentration of ammonia be $C(\ce{NH3})$, then because $\ce{NH3}$ is such a weak base:


$$C(\ce{NH3})\approx [\ce{NH3}]$$


So that we have:


$$K\_b\approx\frac{[\ce{OH-}]^2}{C(\ce{NH3})}$$


And with $pH+pOH=14$ and $[\ce{OH-}]=10^{-pOH}$, you can then calculate $C(\ce{NH3})$, in the $20\ \mathrm{ml}$.


Then calculate the number of moles of $\ce{NH3}$ contained in that volume and calculate how much of the strong $\ce{NH3}$ solution you need to add to the $20\ \mathrm{ml}$ to reach $pH=12.8$.


Use the rule $C\_1(\ce{NH3})\times V\_1=C\_2(\ce{NH3})\times V\_2$ for the above.


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