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

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.

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    $\begingroup$ Please explain/show what attempts you have made to solve this problem on your own; these types of open questions do not to adhere to the posting policy of the site. $\endgroup$ – J. Ari Jan 3 '18 at 19:03
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    $\begingroup$ Sure I will edit the post with what I tried. This is not a homework problem, I am a researcher trying to do chemistry in the lab, but my background is in physics, so I am struggling with some of the elementary concepts. $\endgroup$ – Joey Benson Jan 3 '18 at 19:33
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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.

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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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  • $\begingroup$ This is pretty much what OP has already shown in the question. Have you actually tried to calculate the concentration? It's still going to be two-hundred-something mole per liter; it's not the math that it wrong, it's the initial assumption. $\endgroup$ – andselisk Jan 4 '18 at 0:57

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