Equation for a solution of a weak acid, HA, and its salt NaA...?
derive the Henderson-Hasselbalch equation for a solution of
a weak acid, HA, and its salt NaA...
I have an answer for a solution of a weak acid HA:|
pH = pKa + log10([A-]/[HA])
But the problem is , when the salt NaA (sodium acid) is included, I don't know how to derive it...
Need help anyone? :(
Answer:
NaA is not sodium acid but is a salt
We consider the solution made by adding HA and NaA.
NaA is a strong electrolyte and it will be dissociated 100% into Na+ and A-
The equilibrium is
HA <> H+ + A-
at equilibrium
[HA] = initial molarity HA - x
[ H+] = x
[ A-] = x + initial molarity NaA
K = [H+] [A-] / [HA]
x is small enought to be neglected when added or subctracted from initial molarities
[H+] = K x [HA] / [A-]
log [H+] = log K + log [HA]/ [A-]
- log [H+] = - log K - log [HA]/[A-]
pH = pK + log [A-] / [HA]
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a weak acid, HA, and its salt NaA...
I have an answer for a solution of a weak acid HA:|
pH = pKa + log10([A-]/[HA])
But the problem is , when the salt NaA (sodium acid) is included, I don't know how to derive it...
Need help anyone? :(
Answer:
NaA is not sodium acid but is a salt
We consider the solution made by adding HA and NaA.
NaA is a strong electrolyte and it will be dissociated 100% into Na+ and A-
The equilibrium is
HA <> H+ + A-
at equilibrium
[HA] = initial molarity HA - x
[ H+] = x
[ A-] = x + initial molarity NaA
K = [H+] [A-] / [HA]
x is small enought to be neglected when added or subctracted from initial molarities
[H+] = K x [HA] / [A-]
log [H+] = log K + log [HA]/ [A-]
- log [H+] = - log K - log [HA]/[A-]
pH = pK + log [A-] / [HA]
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