Answer:
Explanation:
Hello,
In this case, since the pH defines the concentration of hydrogen:
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
![[H^+]=10^{-pH}=10^{-3.4}=3.98x10^{-4}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-3.4%7D%3D3.98x10%5E%7B-4%7D)
And the percent ionization is:
![\% \ ionization=\frac{[H^+]}{[HA]}*100\%](https://tex.z-dn.net/?f=%5C%25%20%5C%20ionization%3D%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5BHA%5D%7D%2A100%5C%25)
We compute the concentration of the acid, HA:
![[HA]=\frac{[H^+]}{\% \ ionization}*100\%=\frac{3.98x10^{-4}}{66\%} *100\%\\\\](https://tex.z-dn.net/?f=%5BHA%5D%3D%5Cfrac%7B%5BH%5E%2B%5D%7D%7B%5C%25%20%5C%20ionization%7D%2A100%5C%25%3D%5Cfrac%7B3.98x10%5E%7B-4%7D%7D%7B66%5C%25%7D%20%20%2A100%5C%25%5C%5C%5C%5C)
![[HA]=6.03x10^{-4}](https://tex.z-dn.net/?f=%5BHA%5D%3D6.03x10%5E%7B-4%7D)
Thus, the Ka is:
![Ka=\frac{[H^+][A^-]}{[HA]}=\frac{3.98x10^{-4}*3.98x10^{-4}}{6.03x10^{-4}}\\ \\Ka=2.63x10^{-4}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D%3D%5Cfrac%7B3.98x10%5E%7B-4%7D%2A3.98x10%5E%7B-4%7D%7D%7B6.03x10%5E%7B-4%7D%7D%5C%5C%20%20%5C%5CKa%3D2.63x10%5E%7B-4%7D)
So the pKa is:
Regards.
<span>N2 + 3H2 → 2 </span>NH3<span> from bal. rxn., 2 moles of </span>NH3<span> are formed per 3 moles of </span>H2, 2:3 moleH2<span>: 3.64 </span>g<span>/ 2 </span>g<span>/mole </span>H2<span>= 1.82 1.82 moles </span>H2<span> x 2/3 x 17
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