Ka = (products)/(reactants)
with coefficients becoming exponents.
What are some of the equilibrium expressions that chemists use for weak acid equilibrium?
3 answers
Are there any more expressions for weak acids equilibrium? It would be helpful in the long run to know what they all are if there are any more.
You're question is so general that I didn't know what to write the first time and certainly I don't know what to write now. For a weak acid, such as HA, it ionizes as
HA(aq) + H2O(l) ==> H3O^+(aq) + A^-(aq) in which aq stands for aqueous (as in solution) and l stands for liquid (meaning pure H2O), the
Ka = (H3O^+)(A^-)/(HA)
All follow that same generic Ka. There are some diprotic and triprotic acids but they follow the same rules. For example, H2CO3 is
H2CO3(aq) + H2O(l) ==>H3O^+(aq) + HCO3^-(aq) and k1 for that ionization is
k1 = (H3O^+)(HCO3^-)/(H2CO3)
The second hydrogen then ionizes as
HCO3^-(aq) + H2O(l) ==>H3O^+(aq) + CO3^2-(aq) for which k2 is
k2 = (H3O^+)(CO3^2-)/(HCO3^-)
etc.
H3PO4, a triprotic acid, works the same way and it has a k1, a k2, and a k3 in which one H comes off at a time.
If you have a specific problem it would be easier to answer that than trying to stay with general cases.
HA(aq) + H2O(l) ==> H3O^+(aq) + A^-(aq) in which aq stands for aqueous (as in solution) and l stands for liquid (meaning pure H2O), the
Ka = (H3O^+)(A^-)/(HA)
All follow that same generic Ka. There are some diprotic and triprotic acids but they follow the same rules. For example, H2CO3 is
H2CO3(aq) + H2O(l) ==>H3O^+(aq) + HCO3^-(aq) and k1 for that ionization is
k1 = (H3O^+)(HCO3^-)/(H2CO3)
The second hydrogen then ionizes as
HCO3^-(aq) + H2O(l) ==>H3O^+(aq) + CO3^2-(aq) for which k2 is
k2 = (H3O^+)(CO3^2-)/(HCO3^-)
etc.
H3PO4, a triprotic acid, works the same way and it has a k1, a k2, and a k3 in which one H comes off at a time.
If you have a specific problem it would be easier to answer that than trying to stay with general cases.