ASAP - what is a way to find Ka through titration. - using only mod 5 knowledge - not the henderson hasselbalch equation (1 Viewer)

[amdspotter]

^^^

 

[amdspotter]

^^^

i have a way to do this in mind, using ph at equivalance point then using the formula -ph = log(h+) and solving for H+, then using ice table and using Ka formula
but i dont understand why we should use this at the equivalence point.
can any1 explain
@CM_Tutor ur help here is appreciated.

 

[CM_Tutor]

Plot the pH curve. If V_equiv is the volume required to reach the equivalence point, then at 0.5V_equiv, pH = pK_a.

 

[amdspotter]

Plot the pH curve. If V_equiv is the volume required to reach the equivalence point, then at 0.5V_equiv, pH = pK_a.

But then this is using the Henderson hasselbalch EQ isn't it?

 

[CM_Tutor]

But then this is using the Henderson hasselbalch EQ isn't it?

How?

At the half equivalence point, half of the acid or base has been converted to its conjugate, so the ratio of the two forms is 1:1. Put this into the formula for K_a and the two conjugate forms cancel, leaving K_a = [H₃O⁺].

 

[amdspotter]

How?

At the half equivalence point, half of the acid or base has been converted to its conjugate, so the ratio of the two forms is 1:1. Put this into the formula for K_a and the two conjugate forms cancel, leaving K_a = [H₃O⁺].

Ah ok this makes sense yep thanks a lot