Consider that you have a solution of a weak acid - CH3COOH: 25mL of 0.5M solution
And you have another solution of NaOH = also 25mL of 0.5M NaOH
NaOH reacts with CH3COOH in 1:1 mol ratio:
NaOH + CH3COOH → CH3COONa + H2O
If you mix these two solutions totally, they exactly neutralise themselves and you have 50mL of a CH3COONa solution. Being the salt of a strong base and a weak acid, the solution will be basic, pH > 7.00. There is no unreacted acid and no unreacted base left in the solution.
The above scenario describes the EQUIVALENCE point.
But an interesting thing occurs at what is called the half equivalence point: This is when the 25mL of 0.5M CH3COOH is reacted with only 12.5mL of the NaOH solution. The acid is half neutralised, so there is half the acid remaining un-neutralised
In the original solutions you have:
In the 25mL of 0.5M CH3COOH you have : 25/1000*0.5 = 0.0125 mol CH3COOH
Like wise in the 25mL of 0.5M NaOH you have 0.0125 mol NaOH
Now you add 12.5mL of the NaOH to the 25mL of acid. What happens? You have added 0.0125/2 = 0.00625 mol of NaOH wich reacts with 0.00625 mol CH3COOH to produce 0.00625 mol CH3COONa.
Note that only half the acid has been neutralised.
Therefore in the solution you have 0.00625mol unreacted CH3COOH and 0.00625mol CH3COONa in a total volume of 25+12.5 = 37.5 mL of solution.
Calculate the molarity of these components:
CH3COOH = 0.00625 mol in 0.0375L = 0.1667M
CH3COONa = 0.00625mol / 0.0375L = 0.1667M
What you have produced in reality is an acetic acid/ sodium acetate buffer. The pH of such a buffer is calculated using the Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
The pKa of acetic acid = 4.74.
Therefore the pH of the solution =
pH = 4.74 + log ( 0.1667/0.1667)
pH = 4.74 + log 1.0
pH = 4.74 + 0
pH = 4.74
So: to summarise: at the half equivalence point, the pH of the solution is equal to the pKa of the acid. Which is what you asked for, and is explained.