If the ph at the half-titration point of a monoprotic weak acid is 4. 2, what is the ka of the acid?.

You know that at the equivalence point, the strong base will totally neutralize the weak acid if you're titrating a weak monoprotic acid, which I'll refer to as HA.

HA(aq)+OH−(aq)→A−(aq)+H2O(l)

Therefore, upon adding an equal number of moles of a weak acid and strong base, all of the weak acid's moles will be consumed, leaving you with A, the weak acid's conjugate base.

At this point, you have added enough moles of the strong base to neutralize half of the weak acid molecules in the solution. This is known as the half equivalence point.

The weak acid, the strong base, and the conjugate base are all in 1:1 mole ratios, indicating that what you consume from the weak acid and the strong base, you make as the conjugate base. The reaction will use half of the moles of the weak acid and produce just as many moles of the conjugate base.

As a result, the solution will contain an equal number of moles of the weak acid and its conjugate base at the half equivalence point, indicating that you are now working with a buffer solution.

As you are aware, the Henderson-Hasselbalch equation pH=pKa+log can be used to calculate the pH of a weak acid-conjugate base buffer ([conjugate base][weak acid])

You have the half-equivalence point when

[HA]=[A−]

it suggests that

log([HA][A−])=log(1)=0

As a result, it can be said that the pH of the solution and the pKa of the weak acid are equivalent at the half-equivalence point.

At the halfway point of equivalence: pH=pKa

The acid dissociation constant of the weak acid, Ka, determines the pKa. pKa=log(Ka), which indicates that Ka=10pKa.

Ka=10pH will be present when the two points are half equal.

Enter your value to determine Ka=105.67=2.1106.

To learn more about the half-equivalence point click here-

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