Question:
Does pH = PKa at half equivalence or equivalence and explain your answer?
anonymous
2011-06-14 10:48:17 UTC
What is the difference between equivalence and half equivalence?
Four answers:
Trevor H
2011-06-14 11:58:57 UTC
Consider that you have a solution of a weak acid - CH3COOH: 25mL of 0.5M solution

And you have another solution of NaOH = also 25mL of 0.5M NaOH



NaOH reacts with CH3COOH in 1:1 mol ratio:

NaOH + CH3COOH → CH3COONa + H2O



If you mix these two solutions totally, they exactly neutralise themselves and you have 50mL of a CH3COONa solution. Being the salt of a strong base and a weak acid, the solution will be basic, pH > 7.00. There is no unreacted acid and no unreacted base left in the solution.

The above scenario describes the EQUIVALENCE point.



But an interesting thing occurs at what is called the half equivalence point: This is when the 25mL of 0.5M CH3COOH is reacted with only 12.5mL of the NaOH solution. The acid is half neutralised, so there is half the acid remaining un-neutralised



In the original solutions you have:

In the 25mL of 0.5M CH3COOH you have : 25/1000*0.5 = 0.0125 mol CH3COOH

Like wise in the 25mL of 0.5M NaOH you have 0.0125 mol NaOH



Now you add 12.5mL of the NaOH to the 25mL of acid. What happens? You have added 0.0125/2 = 0.00625 mol of NaOH wich reacts with 0.00625 mol CH3COOH to produce 0.00625 mol CH3COONa.

Note that only half the acid has been neutralised.

Therefore in the solution you have 0.00625mol unreacted CH3COOH and 0.00625mol CH3COONa in a total volume of 25+12.5 = 37.5 mL of solution.

Calculate the molarity of these components:

CH3COOH = 0.00625 mol in 0.0375L = 0.1667M

CH3COONa = 0.00625mol / 0.0375L = 0.1667M



What you have produced in reality is an acetic acid/ sodium acetate buffer. The pH of such a buffer is calculated using the Henderson - Hasselbalch equation:

pH = pKa + log ([salt]/[acid])

The pKa of acetic acid = 4.74.

Therefore the pH of the solution =

pH = 4.74 + log ( 0.1667/0.1667)

pH = 4.74 + log 1.0

pH = 4.74 + 0

pH = 4.74



So: to summarise: at the half equivalence point, the pH of the solution is equal to the pKa of the acid. Which is what you asked for, and is explained.
Aaron
2011-06-14 12:29:15 UTC
There are too many wrong answers here so I feel like Trevor H needs some backup.



I don't see how anyone can use the Henderson-Hasselbach equation to derive this, and STILL get it wrong.



This is the Henderson-Hasselbach equation



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



At half-equivalence point, exactly half of your acid is dissociated by addition of titrant. Therefore, [A-]=[HA].



Therefore, log([A-]/[HA]) = 0



Ergo, pKa = pH at half-equivalence point.



It's unwise to call someone a donkey without checking your facts, Martin. It only makes you look like a bigger idiot.
anonymous
2011-06-14 11:37:22 UTC
pH=log (A-/HA) + pka



HA + BOH --> H2O + BA which is like a general equation representing how an acid and a base combine to make water and a salt.



at half way to the equivalence point the concentrations of A- and HA, which is like the anion of the acid and the strong acid are equal. this is because you have added half the amount of base needed to neutralize the acid. so half the acid is still there and half has been used to make the neutral salt which dissociates into B+ and A-.

so the fraction in the parentheses is equal to one. log(1) is 0. so ph= pka at equivalence point.



for ex. if you have 2.0 M of acid and 2.0 M of base.

if you have 1L of acid you need 1L of base to neutralize all the acid at the equivalence point. So half way to the equivalence point you have 0.5 L of base.

add together.

that's 2.0 moles of acid + 1.0 moles of base.

which means they react to have 1.0 moles of acid left and 1.0 moles of salt. that's...

pH= log (1/1) +pka

pH= 0 +pka

pH = pka



at the equivalence point. the acid has been completely neutralized. Using stoichiometry the moles of acid and base added are the same and cancel and all you have are a neutral salt and water. which means you do a neutral salt calculation using an ICE table.
Shodai Satori
2011-06-14 11:21:21 UTC
pH = -log[H+]



pKa = -logKa



when pH = pKa



[H+] = Ka



for a monobasic acid



Ka = [H+][A-]/[HA] = [H+]^2/[HA]



if Ka = [H+] then [H+]^2/[HA] = [H+]



[H+] = [HA] which is the equivalence point



equivalence point is the point where enough acid has been added to enough base for neutralization in accordance with the stoichiometric equation for the neutralization. Half equivalence is when half the amount of acid/base has been added


This content was originally posted on Y! Answers, a Q&A website that shut down in 2021.
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