Lakhasly

If we titrate the weak acid HB with a base, there will be a point in the titration at which the number of moles of base added is half the number of moles of acid initially presents.From the resultant titration curve we obtain the ionization constant, as explained in the following paragraph.From the titration curve shown in Figure 4.1, we see at the point denoted as half equivalence point, where [HB] = [B?], the pH is 4.3.

If we titrate the weak acid HB with a base, there will be a point in the titration at
which the number of moles of base added is half the number of moles of acid
initially presents. This is the point at which 50% of the acid has been titrated to
produce B‾ and 50% remains as HB. At this point [HB] = [B‾], the ratio [HB]/[B‾]
= 1, and log [HB]/[B‾] = 0. Hence, at this point in a titration, that is, at half the
equivalence point, Equation [8] becomes
pH = pKa [9]
By titrating a weak acid with a strong base and recording the pH versus the
volume of the base added, we can determine the ionization constant of the weak
acid. From the resultant titration curve we obtain the ionization constant, as
explained in the following paragraph.
From the titration curve shown in Figure 4.1, we see at the point denoted as half
equivalence point, where [HB] = [B‾], the pH is 4.3. Thus from Equation [9], at
this point pH = pKa = 4.3, or Ka = 510‾5
.