Hi, I need assistance with the following questions:
1) If the acid dissociation constant, Ksuba, for an acid HA is 8 x 10^-4 at 25 degree C, what percent of the acid is dissociated in a 0.50 M solution of HA at 25 degree C?
2) Equal numbers of moles of Hsub2(g), Ar(g), and Nsub2(g) are placed in a glass vessel at room temperature. If the vessel has a pin sized leak, which of the following will be true regarding the relative values of the partial pressures of the gases remaining in the vessel after some of the gas mixture has effused?
a) PsubH2 < PsubN2 < PsubAr
b) PsubH2 < PsubAr < PsubN2
c) PsubN2 < PsubAr < PsubH2
d) PsubAr < PsubH2 < PsubN2
e) PsubH2 = PsubAr = PsubN2
1) The reaction is:
HA + H2O -> H30^+ + A^- and the concentration changes will be:
0.5 -X X X
- The Ka is calculated as Ka =([H3O]*[A])/[HA]
Thus we can write 8.10^-4 = ...
This solution is comprised of a concise and clear explanation which answers the two chemistry-based questions. An explanation of how to solve the first question is provided, along with a description explaining why the correct option for the second question is in fact correct.