Question

that is he change in when mL solution of NaoH is added to 1.00 L of a solution 1.00 acetate and 1.00 Min acetic acid.

pH change =

Answer 1

The data in the question is incomplete and the question is unclear.

Assuming the problem statement as below:

What is the change in pH when ‘x’ L of ‘y’M NaOH solution is added to 1 L of a buffer solution which is 1 M in sodium acetate and 1 M in acetic acid.

Using the equation:

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

Here initially [salt] = [acid] = 1 M, so pH = pKa = 4.75 (for acetic acid)

Initial moles of salt = molarity * volume = 1*1 = 1 mole

Similarly initial moles of acid = 1 mole

Moles of NaOH added = x*y

After this addition, moles of acid decrease and moles of salt increase by the same amount

Above equation can also be written as :

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

So after adding NaOH, equation becomes:

pH = 4.75 + log( (1+x*y) / (1- x*y) ) = z (say)

Thus the required answer is z-4.75