pOH = pKb at the half-equivelence point?

[Lunasly]

I'm reading about titrations and I have a few questions regarding what is true at the half-equivelence point.

If I am titrating a weak acid with a strong base, then...

1) moles of WA = moles of conjugate base.
2) pH of the solution = pKa of the weak acid.
3) pOH of solution = pKb of the conjugate base.

If I am titrating a weak base with a strong acid, then...

1) moles of weak base = moles of conjugate acid
2) pOH of solution = pKb of weak base.
3) pH of solution = pKa of conjugate acid.

Can anyone please confirm if this is correct? Also, does anyone know how the [H+] relates to the [OH-] at the half equivalence point?

Thank you!

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[Lunasly]

Not sure if we're allowed to bump threads, but any quick confirmation would be greatly appreciated. Thank you for the help!

 

[gettheleadout]

I'm reading about titrations and I have a few questions regarding what is true at the half-equivelence point.

If I am titrating a weak acid with a strong base, then...

1) moles of WA = moles of conjugate base.
2) pH of the solution = pKa of the weak acid.
3) pOH of solution = pKb of the conjugate base.
If I am titrating a weak base with a strong acid, then...

1) moles of weak base = moles of conjugate acid
2) pOH of solution = pKb of weak base.
3) pH of solution = pKa of conjugate acid.

Can anyone please confirm if this is correct?

This is all correct.

Also, does anyone know how the [H+] relates to the [OH-] at the half equivalence point?

Thank you!

The relationship between [H+] and [OH–] depends on the nature and strength of the analyte (and its conjugate, but they go hand in hand); For titration of weak acids the solution will be basic at the half-eq. point, and the more basic the conjugate is (the lower its pKb) the lower the pOH will be and thus the greater the ratio of [OH–]:[H+]. The reverse is true for a titration of a weak base. That make sense?

 

[Lunasly]

Thank you, gettheleadout. That made perfect sense. I appreciate the help.