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.