Weak Acid Strong Base Reactions: Buffer Solutions

[EdTheHead]

First off how exactly do weak acid strong base reactions work. Let's say I have 1 mole of a weak acid HA in a litre of water and let's say for every 4 undissociated molecules 1 molecules dissociate. So I have about 0.2 moles of H+ ions in this solution then I add some KOH let's say 0.5 moles. I know that there will now be 0.5 moles OH- ions that will snatch all the H+ ions. Once the initial 0.2 moles of H+ ions are converted into water and I'm left with no H+ ions and 0.3 moles of OH- ions what happens? Will the weak acid rapidly dissociate in an attempt to regain its usual equilibrium? In other words will the remaining 0.8 moles of HA rapidly become 6.4 moles of HA and 0.16 H+?

With all that in mind would the HA in this case be buffering the solution by neutralizing any base added but maintaining a specific pH based on its Ka? Would I be right to think that even in this solution the pH will be gradually rising as I'm adding the strong base because although the Ka of the weak acid remains constant the concentration is decreasing therefore there will be less overall H+ ions getting fed into the solution?

 

[Borek]

Will the weak acid rapidly dissociate in an attempt to regain its usual equilibrium?

Something like that. You can assume neutralization is stoichiometric and instantaneous.

In other words will the remaining 0.8 moles of HA rapidly become 6.4 moles of HA and 0.16 H+?

No, 0.8 mole of HA would never become 6.4 moles of HA. Up to this moment there was some logic in your post, but something is terribly wrong here.

With all that in mind would the HA in this case be buffering the solution by neutralizing any base added but maintaining a specific pH based on its Ka? Would I be right to think that even in this solution the pH will be gradually rising as I'm adding the strong base because although the Ka of the weak acid remains constant the concentration is decreasing therefore there will be less overall H+ ions getting fed into the solution?

Ka is constant no matter what is present in the solution. As HA is getting easily neutralized, it consumes added base. pH slowly goes up, as described by the Henderson-Hasselbalch equation.

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methods

 

[EdTheHead]

No, 0.8 mole of HA would never become 6.4 moles of HA. Up to this moment there was some logic in your post, but something is terribly wrong here.

Haha sorry I meant 0.64 moles. As in the 4:1 ratio would mean 0.8 moles of HA dissociates into 0.64 moles of HA and 0.16 moles of its conjugate base.

Ka is constant no matter what is present in the solution. As HA is getting easily neutralized, it consumes added base. pH slowly goes up, as described by the Henderson-Hasselbalch equation.

I find the henderson-hasselbach equation very hard to visualize. The log of the fraction part throws me off. If I was to graph a weak acid strong base titration and plot the pH on the y-axis and amount of base added on the x-axis would I see a slow rise in pH until the amount of base added equals the initial amount of weak acid added (in other words when base has completely neutralized the acid) then the pH would begin to rise rapidly?

 

[Borek]

(taken from titrations.info/acid-base-titration-end-point-detection)

This is acetic acid titrated with NaOH. You start with a sharp rise at the very beginning, followed by a flat part (described by Henderson-Hasselbalch equation). Once the buffering effect ends, pH skyrockets.

Yellow-blue band is about indicator color change (thymol blue).