Finding acid dissociation rate constant from titration

[Kavorka]

Homework Statement

When a weak acid HA is titrated 75% of its endpoint, it has a pH of 5.25. Calculate its Ka.

Homework Equations

Ka = [H+ ][A- ]/[HA]
[H+] = 10^-pH

The Attempt at a Solution

I found [H+ ] from the pH to be 10^-5.25 M. From the dissociation equation of HA, we know there is a one-to-one molar ratio between [H+ ] and [A- ], so [A- ] = 10^-5.25. We know that 75% of the original HA has been titrated with the reaction: HA(aq) + OH-(aq) <------> A-(aq) + H2O (l), so there is 3 times the amount of A- than HA (75%:25%).
Ka = [10^-5.25][10^-5.25] / [10-^5.25 / 3] Ka = 1.69x10^-5

Im just not sure if I'm doing this right

 

[epenguin]

Homework Statement

When a weak acid HA is titrated 75% of its endpoint, it has a pH of 5.25. Calculate its Ka.

Homework Equations

Ka = [H+ ][A- ]/[HA]
[H+] = 10^-pH

The Attempt at a Solution

I found [H+ ] from the pH to be 10^-5.25 M. From the dissociation equation of HA, we know there is a one-to-one molar ratio between [H+ ] and [A- ], so [A- ] = 10^-5.25. We know that 75% of the original HA has been titrated with the reaction: HA(aq) + OH-(aq) <------> A-(aq) + H2O (l), so there is 3 times the amount of A- than HA (75%:25%).
Ka = [10^-5.25][10^-5.25] / [10-^5.25 / 3] Ka = 1.69x10^-5

Im just not sure if I'm doing this right

It looks right to me You could have simply said that the ratio[A- ]/[HA] is 3, so K_a = 3×10^-5.25.

It would be better to put in more arithmetical steps – then if you make a mistake you still get nearly full credit.

The phrase I have bolded is definitely not true, and contradicts the rest of what you say.. Also this is nothing to do with rate constants.