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-]
pH = 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.
Calculate the pH of a 0.010 M solution of Acetic Acid
The equation is:
pH = pKa - log[AH]/[A-]
where
- [AH] is the acid concentration
- [A-] is the conjugate base concentration
pH = pKa + log[A]/[HA]
Yes.
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
5.6x10-10 at pH = 8.5
pOH is the negative log of the OH- concentration. It is also related to pH by pH + pOH = 14.
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
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
pH + pOH =14
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
5.6x10-10 at pH = 8.5
what do you mean by 50 KA for 1 sec
pOH is the negative log of the OH- concentration. It is also related to pH by pH + pOH = 14.
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
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
pH= -log = 1.59
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
H2S >> H+ + HS- Ka = [H+][HS-]/[H2S] Ka = 10^-7.0
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
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]