Acid Dissociation Constant?

Question

Please help! There are 3 Q:
1. From the definition of Ka for a weak acid, derive the following equation.
pH= pKa + log10 ([conjugate base]/[acid])
2. Based on the equation from question 1., what is the pH when 50% of an acetic acid was neutralised by sodium hydroxide? (call this pH50%). Similarly, what is the pH when there was a 55% neutralisation? (call this pH55%). What is the expected change between pH55% and pH50%?
3. What would be the final pH if 50 mL of water were added to 50mL of acetic acid/acetate buffer solution initially at pH 5.20? (consider the equation given in question 1.)

Could you please show the steps/workings as well!
THANK YOU!!!

Answer 1

Consider a weak acid dissociating: HA <--> H+ + A-

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

Take the negative log of both sides:

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

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

2. If the acid is half neutralized, the pH = pKa. This is because the ratio [A-]/[HA] = 1 and log 1 = 0. Look up the pKa of acetic acid and you have it.

For the second part, if the acid is 55% neutralized, then [A-] = 0.55 and [HA]= 0.45.

Plug those into the equation and solve for the pH.

3. When you dilute a buffer solution, the pH should not change, because you are not affecting the ratio of A- to HA. You are changing their absolute concentrations, but not the ratio. So, the pH should still be 5.20.

Answer 2

sure you're perfect, so stable acids have an acid dissociating consistent which techniques infinity. consequently Ka is barely commonly used for susceptible acids. If it type of feels wierd undergo in suggestions that Ka is largely an equilibrium consistent however the dissociation of a stable acid isn't possibly an equilibrium technique.

Answer 3

1) Reaction of dissociation of a weak acid is:

Acid + H2O ----------> Conjugate Base + H3O(+)

Definition of Ka:

Ka = [conjugate base][H3O+] / [[Acid]

We apply the operator "p" (= - log) in both sides of equation:

-log(Ka) = -log(H3O+) - log([conjugate base]/[acid])

pKa = pH + log ( [conjugate base]/[acid])

That's it!

Answer 4

1. From the definition of Ka for a weak acid, derive the following equation.
pH= pKa + log10 ([conjugate base]/[acid])

Given association rxn of acetic acid for instance :
CH3COOH (weak acid)<----> (H+) + (CH3COO-)(conjugate base)
Ka = [H+][CH3COO-]/[CH3COOH]

rearranging, [H+] = Ka(CH3COOH)/[CH3COO-]
log on both sides,

log [H+]= log Ka + log (CH3COOH/CH3COO-)

-log[H+]= -log Ka + log (CH3COO-/CH3COOH)

thus pH= pKa + log(conjugate base/acid)


Sorry i am unable to answer Q2 and Q3