If the solution contained pure HPO4 2-, the pH would be an average of pKa2 and pKa3. Why is that? I just thought the pH would depend on pKa3.
Buffer
- Thread starter Sonyfan08
- Start date
It's because two reactions are taking place.
1. HPO4 2- is reacting with H2O to give H2PO4 1- and OH-
2. HPO4 2- is reacting with H2O to give PO4 3- and H3O+
So basically, you have to take into consideration HPO4 2- losing a hydrogen ion in the forward reaction (Pka2) and taking up a hydrogen ion (Pka3) in the reverse reaction.
Hopefully that wasn't too confusing, it's difficult to translate reactions into words 😛
1. HPO4 2- is reacting with H2O to give H2PO4 1- and OH-
2. HPO4 2- is reacting with H2O to give PO4 3- and H3O+
So basically, you have to take into consideration HPO4 2- losing a hydrogen ion in the forward reaction (Pka2) and taking up a hydrogen ion (Pka3) in the reverse reaction.
Hopefully that wasn't too confusing, it's difficult to translate reactions into words 😛
PS 31: If the valve is opened to drain the tank, where is the speed of the flowing water the greatest?
A) At the top of the tank
b) at the bottom of the tank
c) at the wide end of the pipe
d) at the narrow end of the pipe
Ok, so if the flow rate was the same, D would have greater velocity than C. So I eliminated C. From comparing A&B, bournouli equation states that where there is an greater pressure, the velocity is the smaller. So I eliminated B for this reason. I then just picked C over A was because I remembered in physics class that we just assumed the speed at the tank was very small. And going back I can't give a good reason why it's not A.
A) At the top of the tank
b) at the bottom of the tank
c) at the wide end of the pipe
d) at the narrow end of the pipe
Ok, so if the flow rate was the same, D would have greater velocity than C. So I eliminated C. From comparing A&B, bournouli equation states that where there is an greater pressure, the velocity is the smaller. So I eliminated B for this reason. I then just picked C over A was because I remembered in physics class that we just assumed the speed at the tank was very small. And going back I can't give a good reason why it's not A.
- Joined
- Jul 23, 2011
- Messages
- 850
- Reaction score
- 20
if there is a hole in the tank, the hole at the bottom will ahve greater velocity? why? because according to bernoulli, the P term on both side are P (atm) and they are equal. then term pgh of the top valve is higher than the bottom, thus the velocities of the bottom have to be higher than the top. Your reasonning that the top valve having higher velocity only apply to two point of identical height. So, answer B should be picked between A and B.
In the pipe, you can use flow rate equation and find that the narrow part have higher velocities than the wide. so D is picked from C and D.
In the pipe, you can use flow rate equation and find that the narrow part have higher velocities than the wide. so D is picked from C and D.
huh.. can you explain this more. I'm a bit confused. I guess I just always assumed that where there is greater pressure, the velocity is smaller. I thought the pressure was greater at the bottom because of gauge pressure + atm vs the top where there is only atm pressure. I remember the greater presure = smaller velocity because of some problems in TBR with the top and bottom part of the wing.
haha, and sorry for posting in the wrong section. lol, I was supposed to post this in AAMC 8..
- Joined
- Jul 23, 2011
- Messages
- 850
- Reaction score
- 20
Bernoulli equation
P(1) + pgh (1) + .5pv^2 (1) = P(2) + pgh (2) + .5pv^2 (2)
In your situation, there is 2 holes, one on top (1) and one on bottom (2). I assume there point are open to atsmosphere, if they arent and have different pressure, idk how you going to solve em. P (1) = P (2) = Patm, pgh (1) > pgh (2); therefore .5pv^2 (1) < .5pv^2 (2) for the bernouli principle to hold.
When you have 2 point of identical height, you can eliminate the pgh term, your equation become
P(1) + .5pv^2 (1) = P(2) + .5pv^2 (2)
Now, you see which ever one have higher pressure (P term) will have lower .5pv^2 term.
The takeaway fomr this is to write out and equation and apply it to your situation, dont make assumptions.
P(1) + pgh (1) + .5pv^2 (1) = P(2) + pgh (2) + .5pv^2 (2)
In your situation, there is 2 holes, one on top (1) and one on bottom (2). I assume there point are open to atsmosphere, if they arent and have different pressure, idk how you going to solve em. P (1) = P (2) = Patm, pgh (1) > pgh (2); therefore .5pv^2 (1) < .5pv^2 (2) for the bernouli principle to hold.
When you have 2 point of identical height, you can eliminate the pgh term, your equation become
P(1) + .5pv^2 (1) = P(2) + .5pv^2 (2)
Now, you see which ever one have higher pressure (P term) will have lower .5pv^2 term.
The takeaway fomr this is to write out and equation and apply it to your situation, dont make assumptions.
