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How many grams of gaseous ammonia do you need to make $\pu{200 mL}$ aqueous ammonia solution with $\mathrm{pH} = 11.34?$

As the mole ratio is the same for $\ce{H+}$ and $\ce{NH3},$

$$\ce{NH3 + H+ -> NH4+}$$

$$n(\ce{NH3(aq)}) = (\pu{10^{-11.34} mol L^-1})(\pu{0.200 L}) = \pu{9.14E-13 mol}$$

$$m(\ce{NH3(g)}) = (\pu{9.14E-13 mol})(\pu{17.031 g mol^-1}) = \pu{1.6E-11 g}$$

The answer seems way too small. What have I not taken into the account?

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  • $\begingroup$ chemistry.stackexchange.com/questions/139089 $\endgroup$
    – andselisk
    Jan 12 at 16:53
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    $\begingroup$ As the mole ratio is the same for H+ and NH3, - It is not. Not even close. Off by multiple orders. $\endgroup$
    – Poutnik
    Jan 12 at 17:15
  • $\begingroup$ @Poutnik Isn't the ratio for H+ and NH3, 1:1, seen the equation? NH3 + H+⟶ NH4+ $\endgroup$
    – Allistor
    Jan 12 at 17:29
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    $\begingroup$ That is ratio for the reaction, not for the occurence, There is more than $\pu{e10}$ times more NH3 than H+. $\endgroup$
    – Poutnik
    Jan 12 at 17:43
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    $\begingroup$ Amount ratio OH- : NH3 is not 1 : 1 either. You are closer, but still few orders off. You are guessing. // Kb = 10^(-4.75)=[NH4+][OH-]/[NH3] $\endgroup$
    – Poutnik
    Jan 12 at 19:25

1 Answer 1

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This problem is an equilibrium problem, and namely the following equilibrium : $$\ce{NH3 + H2O = NH4^+ + OH-}$$ The tables give the equilibrium constant : $$\ce{$K$_b = \frac{[NH4^+][OH-]}{[NH3]} = 1.8 10^{-5}}$$ Let's define $c$ as the total concentration of $\ce{NH3}$ brought into solution (included the part transformed into $\ce{NH4^+}$), of which the amount $\alpha$ (< $c$) has been reacted and transformed into $\ce{NH4+ and OH-}$. The problem is to find the value of $c$. Now $c$ - $\alpha$ is the residual concentration of $\ce{NH3}$ in solution.

As the final pH is $11.34$, $\pu{[OH^-]}$ = $\alpha$ = $\pu{\frac{10^{-14}}{[H^+]} = 10^{-14 + 11.34} = 10^{-2.66}}$. This $\ce{[OH-]}$ is also equal to $\ce{[NH4^+]}$ Introducing this value in $K_b$ yields : $$\ce{$K$_b = 1.8 10^{-5} = \frac{\alpha^2}{c - \alpha} = \frac{{10^{-5.32}}}{c - 10^{-2.66}}}$$ The value of the concentration $c$ is : $$c - \pu{10^{-2.66} = \frac {10^{-5.32}}{1.8· 10^{-5}} = 0.05555} \mathrm{mol/L} $$ As $\ce{10^{-2.66} = 0.002188}$ it gives : $$c = 0.0577 \mathrm{mol/L}$$ $$n\ce{(NH3) = 0.2 \mathrm{L} · 0.0577} \mathrm {mol/L} = 0.01154 \mathrm{mol}$$ This corresponds to a volume of gaseous $\ce{NH3}$ equal to $$\ce{V(NH3) = \frac{$nRT$}{$P$} = 0.01154\frac{$RT$}{$P$} = 0.282 L}$$

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  • $\begingroup$ I just calculated it myself and saw your comment, I'm very glad it seems to be the same as you have deduced. It's convenient to be able to remove some of my doubts, much appreciated! $\endgroup$
    – Allistor
    Jan 12 at 20:19

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