|
|||||||
| DAT Discussions Discuss study tips and resources to excel on the Dental Admissions Test. | RSS: |
![]() |
|
|
Thread Tools | Display Modes |
|
|
#1 |
|
Senior Member
|
SDN Members don't see this ad. (About Ads)
|
|
|
|
|
|
#2 |
|
1K Member
|
pKa = -log of [Ka]
pH = -log of [H+] Ka is the acid dissociation constant = [H+][A-]/[HA] pH= pKa when the conjugate base = weak acid at the half-way point on a titration curve the lower the pKa the more acidic the solution is the lower the pH the more acidic the solution is |
|
|
|
|
|
#3 |
|
Senior Member
|
thanks, so if in a problem i were given the pKa, how would i convert that to pH?
|
|
|
|
|
|
#4 |
|
Senior Member
|
They would also have to give you the concentration of the acid and its conjugate base so you could use the henderson hasselbach formula. Otherwise, it would be pretty impossible to derive the pKa. You can however tell whether the acid is strong or weak, etc.
|
|
|
|
|
|
#5 |
|
1K Member
|
u can use the henderson-hasselbach equation
pH= pKa + log [conj base]/[wa] im not sure of others but hopefully someone else knows. |
|
|
|
|
|
#6 |
|
3K Member
|
You can compare the two, but they are not the same.
They are equal only if [HA] = [A-] |
|
|
|
![]() |
| Bookmarks |
«
Previous Thread
|
Next Thread
»
| Thread Tools | |
| Display Modes | |
|
|
All times are GMT -7. The time now is 02:26 AM.










Linear Mode

