1 (A) This relation gives the dissociation constant of acetic acid (CH3COOH)

Ka = [H3O+][CH3COO-]\[CH3COOH]

Starting from this relationship show that

pH = pKa + log [base]\[acid]

urgently need assistance

Start with the acid dissociation equation:

HAc -> H+ + Ac-

Ka is then

[H+][Ac-]/[HAc] {given in the question}

at the start

HAc -> H+ + Ac-

and we start from A mole/L and B mol/L for the acid and buffer

A M ___0__ B M

at the end

HAc -> H+ + Ac-
A-x____x___B+x

so Ka is
Ka=(x)(B+x)/(A-x)

Assume that x is small wrt A and wrt B so that the equation reduces to

Ka=x(B)/(A)

which we can rearrange to

x=Ka(A)/(B)

take logs of each side

log(x)=log(Ka(A)/(B))

which is

log(x)=log(Ka) + log((A)/(B))

multiply through by -1

-log(x)=-log(Ka) - log((A)/(B))

which is

-log(x)=-log(Ka) + log((B)/(A))

or

pH = pKa + log((B)/(A))

or

pH = pKa + log [base]\[acid]

As an added note (perhaps picky but I think needed) is that / is used as a divide sign while \ is used for other purposes.

Sure, I can help you with that!

To start, let's take the equation for the dissociation constant of acetic acid and take the logarithm of both sides:

Ka = [H3O+][CH3COO-]/[CH3COOH]

Taking the logarithm of both sides:

log(Ka) = log([H3O+][CH3COO-]/[CH3COOH])

According to the properties of logarithms, we can split the numerator:

log(Ka) = log([H3O+]) + log([CH3COO-])/[CH3COOH]

Now, let's use the relationship between pKa and Ka. The pKa is defined as the negative logarithm of the Ka:

pKa = -log(Ka)

Substituting this into our equation:

-log(Ka) = log([H3O+]) + log([CH3COO-])/[CH3COOH]

Multiplying both sides by -1, the negative sign cancels out:

log(Ka) = -log([H3O+]) - log([CH3COO-])/[CH3COOH]

Now, let's use the relationship between pH and [H3O+]. pH is defined as the negative logarithm of [H3O+]:

pH = -log([H3O+])

Substituting this into our equation and rearranging the terms:

pH = -log([H3O+]) - log([CH3COO-])/[CH3COOH]

Since [H3O+] is the concentration of the acid and [CH3COO-] is the concentration of the base, the equation can be written as:

pH = pKa - log([base]/[acid])

And that's the desired result: pH = pKa + log([base]/[acid]).

This equation relates the pH of a solution to the pKa (the negative logarithm of the dissociation constant) and the ratio of the concentrations of the conjugate base ([base]) to the acid ([acid]).

Sure! I can help you derive the relationship between pH, pKa, and the ratio of base to acid using the given dissociation constant (Ka) equation for acetic acid.

Step 1: Write the expression for the dissociation constant (Ka) of acetic acid:
Ka = [H3O+][CH3COO-]/[CH3COOH]

Step 2: Take the logarithm on both sides of the equation and rearrange:
log(Ka) = log([H3O+][CH3COO-]/[CH3COOH])

Step 3: Use the logarithmic properties to separate the terms:
log(Ka) = log([H3O+]) + log([CH3COO-]) - log([CH3COOH])

Step 4: Rewrite the logarithms with the pH and pKa expressions:
log([H3O+]) = -pH
log(Ka) = -log([CH3COOH]) + log([CH3COO-]) = -log([acid]) + log([base])
log(Ka) = -log([acid]) + log([base])

Step 5: Substitute the expressions from Step 4 into the equation in Step 3:
-log([acid]) + log([base]) = -pH

Step 6: Rearrange the equation to isolate the pH:
pH = log([acid]) - log([base])

Step 7: Finally, apply the logarithmic property for subtraction (log(a) - log(b) = log(a/b)):
pH = log([acid]/[base])

Since pKa is defined as the negative logarithm of the acid dissociation constant (pKa = -log(Ka)), we can substitute -log(Ka) for log([acid]) in the equation:
pH = pKa + log([base]/[acid])

The derived equation is pH = pKa + log([base]/[acid]).