<span>Answer:
The HCl and KOH will react until one or the other is gone. As you have a larger volume of an equal concentration of HCl, the KOH will go first.
moles HCl = 0.04000 L * 0.100 M = 0.00400 moles
moles KOH = 0.02500 L * 0.100 M = 0.00250 moles
moles HCl left = 0.00400 - 0.00250 = 0.00150 moles
Your total volume is now 65.00 mL, so the [HCl] = 0.00150 moles / 0.06500 L = 0.0231 M = [H+]
pH = -log [H+] = -log (0.0231) = 1.64</span>
Explanation:
Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.
Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.
No. of moles =
= 
= 0.2 mol
Molarity of acetic acid is calculated as follows.
Density = 
1 g/ml = 
volume = 100 ml
Hence, molarity = 
= 
= 2 mol/l
As reaction equation for the given reaction is as follows.

So, moles of NaOH = moles of acetic acid
Let us suppose that moles of NaOH are "x".
(as 1 L = 1000 ml)
x = 20 L
Thus, we can conclude that volume of NaOH required is 20 ml.
The is the B) doing background research about the question