Mass of gaseous ammonia required to prepare ammonia solution of a given pH

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?

• chemistry.stackexchange.com/questions/139089 Jan 12 at 16:53
• As the mole ratio is the same for H+ and NH3, - It is not. Not even close. Off by multiple orders. Jan 12 at 17:15
• @Poutnik Isn't the ratio for H+ and NH3, 1:1, seen the equation? NH3 + H+⟶ NH4+ Jan 12 at 17:29
• That is ratio for the reaction, not for the occurence, There is more than $\pu{e10}$ times more NH3 than H+. Jan 12 at 17:43
• 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] Jan 12 at 19:25

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 $$\ce{V(NH3) = \frac{nRT}{P} = 0.01154\frac{RT}{P} = 0.282 L}$$