Calculate the volume of gaseous ammonia $(K_\mathrm{b} = \pu{1.8E-5})$ to dissolve (at NTP) in $\pu{100 mL}$ of water in order to obtain a solution with $\mathrm{pH} = 11.0.$ Consider volume variation negligible.

I started by evaluating molar concentration of $\ce{NH4+}$ and $\ce{OH-}$ knowing the $\mathrm{pH},$ then I used the $K_\mathrm{b}$ inverse formula to evaluate the molar concentration of ammonia. These are the results:

$$c(\ce{NH3}) = \pu{0.056 mol L-1}$$

$$c(\ce{NH4+}) = c(\ce{OH-}) = \pu{1E-3 mol L-1}$$

Then my idea was to work with water. Since $\ce{NH3}$ is a weak base it does not dissociate completely, the amount of water is

$$n(\ce{H2O}) = \frac{\pu{100 g}}{\pu{18.02 g mol-1}} = \pu{5.55 mol}$$

Then I run out of ideas. I tried to find the mamount of ammonia (but I think the result was not correct) by evaluating the degree of dissociation, but then I got lost.

The final result of the problem is $\pu{0.13 L}.$ How do I pursue the problem further?

  • $\begingroup$ I tried also with ideal gas equation. Probably I did some mistake in mole calculation of ammonia moles $\endgroup$ – Davide Gariglio Aug 19 '20 at 8:57
  • $\begingroup$ Probably I solved it. Molarity of ammonia = 0.056 Sinche the volume variation is negligible and 0.056 are the moles of ammonia required to form a liter of solution, the volume of the final solution will be 0.1L. So, the number of moles of ammonia required to have a final volume of 0.1L are: Molarity of ammonia * V final solution = 0.056*0.1 = 0.0056 Putting this number of moles in V=nRT/P the result is 0.1256 Is it correct? $\endgroup$ – Davide Gariglio Aug 19 '20 at 10:03
  • $\begingroup$ Henry constant is not needed. $\endgroup$ – Poutnik Aug 19 '20 at 11:10
  • $\begingroup$ Is my comment correct? $\endgroup$ – Davide Gariglio Aug 19 '20 at 11:43
  • $\begingroup$ If you cannot decide yourself, you do not understand what you have done yet. For numbers to be correct, your algebra must be correct. Show your algebra.( in the question ,not in comments ). Be aware SE Chemistry community is not much interested in validation of particular numerical results, but rather in the Q author understanding of the principles. $\endgroup$ – Poutnik Aug 19 '20 at 12:31

The $\mathrm{pH}$ of your final ammonia solution is $11.0$. When you dissolve ammonia in water it tends to ionize with water to give following equilibrium:

$$\ce{NH3 + H2O <=> NH4+ + OH-} \quad K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = 1.8 \times 10^{-5}$$

Suppose final concentration of $\ce{NH3}$ is $c$, and equlibrium concentrations of $\ce{NH4+}$ and $\ce{NH4+}$ are $\alpha$. Thus:

$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = \frac{\alpha \cdot \alpha}{c - \alpha} = 1.8 \times 10^{-5}$$ $$\alpha^2 + 1.8 \times 10^{-5}\alpha - 1.8 \times 10^{-5}c =0 \tag1$$ Since, $\mathrm{pOH}$ of the solution is $3.0$ ($14 - \mathrm{pH}$), $[\ce{OH-}] = \alpha = 1.00 \times 10^{-3}$, from equation $(1)$:

$$(1.00 \times 10^{-3})^2 + 1.8 \times 10^{-5}\times 1.00 \times 10^{-3} = 1.8 \times 10^{-5}c $$

$$1.8 \times 10^{-5}c = 1.00 \times 10^{-6} + 1.8 \times 10^{-8} = 1.018 \times 10^{-6} $$

$$\therefore \ \ c = \frac{1.018 \times 10^{-6}}{1.8 \times 10^{-5}} = \pu{5.66 \times 10^{-2} molL-1} $$

This is exactly what you got without showing your work. However, you made only $\pu{100 mL}$ of solution. Thus, amount of ammonia dissolved in $\pu{100 mL}$ of water is:

$$$\pu{0.100 L} \times \pu{5.66 \times 10^{-2} molL-1} = \pu{5.66 \times 10^{-3} mol}$

Assuming gaseous ammonia is behaving like ideal gas, you can use $PV = nRT$ here to find the volume of $\pu{5.66 \times 10^{-3} mol}$ of ammonia at STP:

$$V= \frac{nRT}{P} = \frac{(\pu{5.66 \times 10^{-3} mol})(\pu{0.082 LatmK-1mol-1})(\pu{298 K})}{\pu{1 atm}} = \pu{0.138 L}$$


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