Asked Mar 15, 2020

Saccharin, a sugar substitute, is a weak acid with pKa = 2.32 at 25 degrees Celsius. It ionizes in aqueous solution as follows:

HNC7H4SO3 (aq) <=> H+ (aq) + NC7H4SO3- (aq)

What is the pH of a 0.12 M soution of this substance? Express your answer using only 2 decimal places.


Expert Answer

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Step 1

Since pKa = -logKa 

Hence ka = 10-pKa  = 10-2.32

assuming a concentration of Saccharin is dissociating into NC7H4SO3-and H+. 

Hence concentration of NC7H4SO3-and H+ will be = a since 1 mole will produce 1 mole of both.

now since Ka = [NC7H4...

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