# Strong base weak acid titration

1. Aug 9, 2008

### Oerg

Hi All

I have a question that I require an answer to. For the above titration, concentration of A-(conjugate base of acid) is the same as concentration of HA at half equivalence. It is at this point that pka=ph of the solution. I agree to this conclusion, but why is the concentration of A- equals to concentration of HA at this point? Any help will be deeply appreciated. Thanks.

2. Aug 9, 2008

### Staff: Mentor

Write reaction equation, think in terms of the neutralization stoichiometry. This is a weak acid, so it dissociates very slightly on its own - so slightly, that its own dissociation can be ignored.

3. Aug 9, 2008

### Oerg

so, let HA be a weak acid

let x be the amount of acid dissociated. so starting with 1 mol of HA

------------HA <-> H+ + A-
conc--------1-x ----x ---x

in order to achieve the required concentration, conc of HA = conc of A-

1-x=x
x=0.5

0.5 mol of H+ is displaced which requires 0.5 mol of Naoh which is also half of total naoh required, am i right?

4. Aug 9, 2008

### Staff: Mentor

You are mostly right, but you are using ICE instead of following simple stoichiometry.

HA + NaOH -> NaA + H2O

5. Aug 9, 2008

### Renge Ishyo

You can see this pretty easy from the Henderson Hasslebach equation:

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

When [A-] is equal to [HA], you get log 1 in the above equation which is equal to 0. So pH = pKa. The Henderson Hasslebach equation is derived from the standard Ka equilibrium equation, and you can see that derivation if you want a deeper understanding.