Question

Saccharin, a sugar substitute, is a weak acid with pka=2.32 at 25 ∘c. it ionizes in aqueous solution as follows: hnc7h4so3(aq)←−→h+(aq)+nc7h4so−3(aq) part a what is the ph of a 0.11 m solution of this substance?

Answer 1

Saccharin is considered as weak acid:
pH of weak acid = $\frac{1}{2} pKa + \frac{1}{2} pCa$
pKa = 2.32 (given) and
pCa = -log (acid concentration) = - log (0.11) = 0.96
so pH = $(\frac{1}{2}* 2.32 ) + ( \frac{1}{2} * 0.96) = 1.64$