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Dahasolnce [82]
3 years ago
6

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?
Chemistry
1 answer:
MrMuchimi3 years ago
3 0
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
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