Calculate the amount (in mL) of 1.520M NaOH that is required to add the following acetic acid solution to prepare a buffer with the corresponding pH:

30.00mL of a 5.00% (w/v%) acetic acid; the resulting acetate buffer has a pH of 5.75.

pKa of acetic acid = 4.74

So I got this: 5% w/v means 5g HAc/100 mL solution which is (5/60) mols/100 and that is 5/60/0.1L or 0.0833 M.
Then millimols HAc is 30 x 0.0833 = 25.
So acid + base = 25
pH = pKa + log(base)/(acid)
5.75 = 4.74 + log b/a

How do I determine b/a?

equation 2 is a + b = 25

How can I solve for a and b?

You had this in math. It isn't chemistry.

5.75 = 4.74 + log b/a
1.01 = log b/a
10^(b/a) = 1.01
Plug into the calculator; a/b = 10.23

If a/b = 10.23, what is a and what is b?

To determine the values of 'a' and 'b' in the equation a + b = 25, we can use a system of equations.

Since we know that a + b = 25, we can substitute one variable in terms of the other to solve for 'a' or 'b'. Let's solve this equation for 'b':

a + b = 25

b = 25 - a

Now we can substitute this value of 'b' into equation 1 (pH = pKa + log(b/a)):

5.75 = 4.74 + log((25 - a)/a)

To solve this equation for 'a', we need to isolate the logarithmic term. Rearrange the equation:

0.75 = log((25 - a)/a)

Taking the antilogarithm of both sides gives:

10^0.75 = (25 - a)/a

Now, we can solve for 'a' by multiplying both sides by 'a':

10^0.75 * a = 25 - a

10^0.75 * a + a = 25

(10^0.75 + 1) * a = 25

Finally, divide both sides by (10^0.75 + 1) to get the value of 'a':

a = 25 / (10^0.75 + 1)

Once you have the value of 'a', you can substitute it back into the equation b = 25 - a to find the value of 'b'. This will give you the amount (in mL) of 1.520M NaOH required to prepare the buffer solution.