Henderson Hasselbach Equation

[Hemichordate]

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I know the equation is:

pH = pKa + log [A-]/[HA]

But what would the equation be for pOH?

 

[crazybob]

pH = pKa + log([A-]/[HA])

Assuming that you know [A-], [HA], and pKa;

pH + pOH = 14
pOH = 14 - pH = 14 - (pKa + log([A-]/[HA])) = 14 - pKa - log([A-]/[HA]) = 14 - pKa + log([HA]/[A-])

 

[ezsanche]

pH = pKa + log([A-]/[HA])

Assuming that you know [A-], [HA], and pKa;

pH + pOH = 14
pOH = 14 - pH = 14 - (pKa + log([A-]/[HA])) = 14 - pKa - log([A-]/[HA]) = 14 - pKa + log([HA]/[A-])

and since PKa +PKb = 14 and pKb = 14-pka

we get

pKb + log ([HA]/[A-]) =pOH

 

[UCImed]

I think finding pH using the hendersen hasselbach equation and then just subtracting that from 14 is the easiest way 😀

Not a very scientific way but it works

 

[Hemichordate]

I think finding pH using the hendersen hasselbach equation and then just subtracting that from 14 is the easiest way 😀

Not a very scientific way but it works

Well I would do that too, except the question asked for the formula in terms of pOH.

 

[kentavr]

I'd disagree. If they asked about pOH then they probably(?) mean the base reaction, which is:

B + H2O -> HB +OH.

If that is so, then

Kb = [HB][OH] / =>

pKb= - log([HB][OH] / ) = -log([HB]/) - log[OH] = -log([HB]/) + pOH

=>

pOH = pKb+log([HB]/)

 

[thebillsfan]

"and since PKa +PKb = 14 and pKb = 14-pka

we get

pKb + log ([HA]/[A-]) =pOH"

can you explain how that's accurate? i dont see where you go from the first step to the second step

also...EK says youre only supposed to use the HH equation around the buffer region. why CANT you use the HH equation at ANY relative concentration of HA to A-?

thanks everyone

 

[crazybob]

at the buffer region, the pH doesn't change quickly with adding more acid or base. at others, it changes more quickly, and using the HH equation would give you a pH too low or too high.

 

[thebillsfan]

at the buffer region, the pH doesn't change quickly with adding more acid or base. at others, it changes more quickly, and using the HH equation would give you a pH too low or too high.

i realize the pH changes quickly at other regions, but i dont udnerstand why that makes a difference. you have a known concentration of undissociated and dissociated acid. if we assume that those concentrations are not changing, then it shouldnt make a different what region of the titration we're at, right (i'm wrong, obviously, but i dont understand why)

 

[ezsanche]

I'd disagree. If they asked about pOH then they probably(?) mean the base reaction, which is:

B + H2O -> HB +OH.

If that is so, then

Kb = [HB][OH] / =>

pKb= - log([HB][OH] / ) = -log([HB]/) - log[OH] = -log([HB]/) + pOH

=>

pOH = pKb+log([HB]/)

You said the same thing that i said but you replaced my A with a B.

 

[kentavr]

You said the same thing that i said but you replaced my A with a B.

I agree, they look very similar, but there is one significant difference. In my equation the numerator (HB) is a product in your it is a reactant. Also "your A" is product, my "B" is reactant.