PH of a solution obtained by mixing equal volume of 0.2M NaOH & 0.2 Monday CH3CooH (ka=10to the power -5)
Weak acid is mixed with strong base
So the concentration of the salt CH3COONa = 0.1 M
pKa = - log 10-5 = 5
pH= 1/2[pKw - pKa -log C]
= 1/2[14 - 5 -log 0.1]
= 1/2[10]
= 5
So the concentration of the salt CH3COONa = 0.1 M
pKa = - log 10-5 = 5
pH= 1/2[pKw - pKa -log C]
= 1/2[14 - 5 -log 0.1]
= 1/2[10]
= 5