0
Ka = [H+][A-] / [HA]
Hence
[H+] = Ka[HA] / [A-]
Remember pH = -log(10)[H+]
'logging' both sides.
-log(10)[H+] = - log(10)Ka[HA] / [A-]
By algebraic manipulation of log.
pH = -log[A-]^-1 - logKa - log[HA]
pH = log[A-] - logKa - log[HA]
pH = pKa - log[HA]/[A-]
lenpollock
Wiki User
∙ 11y agopH = pKa + log([A-]/[HA])
pH = pKa+log([conjugate base]/[undissociated acid])
pKa is also a measure of the strength of an acid. A low pKa is a strong acid, a higher pKa is a weak acid.
Wiki User
∙ 11y agoCalculate the pH of a 0.010 M solution of Acetic Acid
Anonymous
Wiki User
∙ 8y agoThe equation is:
pH = pKa - log[AH]/[A-]
where
- [AH] is the acid concentration
- [A-] is the conjugate base concentration
Wiki User
∙ 14y agopH = pKa + log[A]/[HA]
Wiki User
∙ 12y agoYes.
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How is the KB of ammonia calculated from the half titration point?
At half titration pH=pKa (you need the pH from the graph of your titration, y axis) ph = pKa + log (base/acid) 10^-pKa = Ka Kw=Ka*Kb Kb=Kw/Ka Ka = Kw/Kb
What is the ka of Ammonia?
5.6x10-10 at pH = 8.5
What is pOH calculation?
pOH is the negative log of the OH- concentration. It is also related to pH by pH + pOH = 14.
What is the pka of .012M solution and a pH of 2.31?
HA ==> H+ + A-Ka = [H+][A-][HA] and from pH = 2.31, calculated [H+] = 4.89x10^-3 M Ka = (4.89x10^-3)(4.89x10^-3)/0.012 Ka = 1.99x10^-3 pKa = 2.70
What would be the pH of a solution when H3PO4 equals H2PO4?
Assuming the Ka= [H+][PO2-]/[PO3-] and that PO3=PO2- then we can safely assume Ka= [H+][PO2-]/[PO2-] and so Ka= [H+][PO2-]/[PO2-] Ka=[H+] since the Ka of Phosphoric acid is equal to 7.5x10-3 then we can take -log(7.5x10-3) to find the pH=2.12
Formula for calculation pH and pOH?
pH + pOH =14
How is the KB of ammonia calculated from the half titration point?
At half titration pH=pKa (you need the pH from the graph of your titration, y axis) ph = pKa + log (base/acid) 10^-pKa = Ka Kw=Ka*Kb Kb=Kw/Ka Ka = Kw/Kb
What is the ka of Ammonia?
5.6x10-10 at pH = 8.5
Busbar fault level calculation?
what do you mean by 50 KA for 1 sec
What is pOH calculation?
pOH is the negative log of the OH- concentration. It is also related to pH by pH + pOH = 14.
What is the pka of a .012M solution and a pH of 2.31?
HA ==> H+ + A-Ka = [H+][A-][HA] and from pH = 2.31, calculated [H+] = 4.89x10^-3 M Ka = (4.89x10^-3)(4.89x10^-3)/0.012 Ka = 1.99x10^-3 pKa = 2.70
What is the pka of .012M solution and a pH of 2.31?
HA ==> H+ + A-Ka = [H+][A-][HA] and from pH = 2.31, calculated [H+] = 4.89x10^-3 M Ka = (4.89x10^-3)(4.89x10^-3)/0.012 Ka = 1.99x10^-3 pKa = 2.70
Calculate the pH of 0.2 M HNO2 solution ( Ka = 4.5 * 10 -5 )?
pH= -log = 1.59
What is Henderson-Hasselbalch equation?
Its an equation you can use to find the pH of a solution. it is.... --- pH = pKa + log (Base/Acid) --- these may help too Ka = 10^-pKa Kw = Ka*Kb
How do you determining pH from weak acids?
H2S >> H+ + HS- Ka = [H+][HS-]/[H2S] Ka = 10^-7.0
What would be the pH of a solution when H3PO4 equals H2PO4?
Assuming the Ka= [H+][PO2-]/[PO3-] and that PO3=PO2- then we can safely assume Ka= [H+][PO2-]/[PO2-] and so Ka= [H+][PO2-]/[PO2-] Ka=[H+] since the Ka of Phosphoric acid is equal to 7.5x10-3 then we can take -log(7.5x10-3) to find the pH=2.12
Why is the PH of the midpoint predicted to be the same value as the weak acid?
if you mean why is it when [H+] = [A-] related to the Ka of a weak acid. look at the definition of Ka. Ka = [H+][A-]/[HA]
