Question

A 40.6 ml sample of a 0.533 M aqueous hypochlorous acid solution is titrated with a 0.349 Maqueous potassium hydroxide solution. What is the pll after 22.9 ml. of base have been added? pH =

Answer 1

Hi there, this is an example for buffer system, unfortunately you have not provided acid dissociation value (Ka) for HClO but i have borrowed this value from sapling website. Please be cautious and cross-check this value from your online course reference values, if matched then enter provided answer, else report it to me below comments section so that i will try my best to respond asap.

HClO Ka = 4.0 x 10^-8

we have to use ICE table

moles HClO = 40.6 mL x 10^-3 L x 0.533 M = 0.0216398 mol

moles KOH = 22.9 mL x 10^-3 L x 0.349 M = 0.0079921 mol

R.................HClO............+.........OH-..............<==>......ClO-..........+.......H2O

I................. 0.0216398.............0.0079921......................................................

C...............-0.0079921.............-0.0079921.................+0.0079921....................

E................. 0.0136477 ...............0..........................+0.0079921....................

pKa = -logKa = -log[4.0 x 10^-8] = 7.39794

pH = pKa + log[conjugate base/weak acid] = 7.39794 + log[0.0079921/0.0136477] = 7.16554144191

pH = 7.16

A 40.6 ml sample of a 0.533 M aqueous hypochlorous acid solution is titrated with a 0.349 Maqueous potassium hydroxide solution. What is the pll after 22.9 ml. of base have been added? pH = 7.16

Hope this helped you!

Thank You So Much!

Please Rate this answer as you wish.("Thumbs Up")