Yes. pH2S = 0.043 but I wouldn't throw away the third place since you are allowed 3 significant figures.
Then Kp = pNH3*pH2S
You know Kp and pH2S; calculate pNH3.
B)XH2S = pH2S/total P
XNH3 = pNH3/total P
C) You have the 0.300 g H2S. From the data above you can calculate the mass NH3 in the 5 L. Add mass NH3 to mass H2S to arrive at the minimum mass of NH4HS.
Thank you for answering me! You don't know how much it means to me. I think I'm over thinking this but I need all the help I can get.
So for A) I just divided .120/.0431 and got 2.78. It just seems too simple to me haha.
For letter C I tried finding the moles of H2S. Since its one to one ratio wouldn't the moles be the same? Then placing these two over Kp to give me the moles of that?
Once again thank you for your response!
No, mols are not the same. They would be if you had the pure NH4HS coming to equilibrium but you don't have that. The problem has charged the vessel with 0.300 g H2S (0.0431 moles H2S); therefore, your value of 2.78 atm is the partial pressure of NH3. Knowing p NH3 you should be able to obtain n from PV = nRT and from that you can get grams NH3.
Oh okay that makes more sense now. Thank you for clearing that up. I tried entering in 2.78 for my partial pressure of NH3 but the program I am using is telling me that is incorrect?
That's my error. The vessel is charged with 0.300 g H2O (0.04315 atm) but that isn't at equilibrium. It will shift to come to equilibrium.
If the system were allowed to come to equilibrium (without the added H2S), the partial pressure of each gas would be sqrt(Kp) = about 0.35 atm. Since the added 0.300 g H2S has a pressure of 0.04315 atm initially, we note that is smaller than the H2S would have been had no H2S been added at all. Therefore, I believe the reaction will shift to the right since there is insufficient H2S.
.........NH4HS ==> NH3 + H2S
Solve for p.
If I didn't make an error my number is 0.3255 for p and 0.3687 for 0.04315+p.
That seems reasonable since
(0.3255*0.3687) = 0.120. Remember I have too many s.f. in my numbers; only 3 s.f. should be used.
Ah! Yes that worked out perfectly! Thank you so much for the help. That one really had me stuck!
Idk why maybe because its 2 am and i cant do things right at this time, but am also trying to solve this problem and when i get to the part to use the quadratic formula to find p i get a different number i have it set up as this:
and i get two answers for p obviously, 1 is .3686 and the other one is -.3255 (which is the same as your answer except is negative) I wonder what am doing wrong >< , thx for all your help
I re-worked the problem and I get two roots for the equation; one is -0.36866 (the same as yours but mine is minus) and +0.3255 (the same as your but mine is +). If you are solving this by hand you are slipping a sign somewhere; if you are using the calculator you are using the negative of the correct formula.
could you please explain the last question, c, again. Ive tried a few times but I am still getting it wrong.
yep that is what happened, i forgot one sign, thank you
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