Question

When 50ml of 0.1M NaOH is added to 50ml of 0.1M CH3COOH solution, the pH will be [pka = 4.74, log5 = 0.7]

Answer 1

Answer:

Explanation:

The answer is simple.

pH= pKa+ log (salt/acid).

In the mixture, No.of m.moles of acetic acid=20 x0.25= 5 m moles.

The no.of m moles of NaOH =50x0.1= 5 m.moles.

So log (salt/acid)= (log1/1) =0

So pH= pKa= 4.74