Question

HCIO2(aq) + H20() Ka 0.012 CIO,-(aq)-H30-(aq) Weak Acid pH 1.54 1 HC102 O CH CICOH O HF C1O2 added no change Equilibrate Reset Time → Initial Equilibrium New Initial [HCIO2] = 0.100 M 7.09 x 10-2 M 7.09 × 10-2 M [H307-0.000 M 2.91 x 10-2 M 2.91 × 10-2 M [Cio21 0.000 M 2.91 10-2 M 2.91 x102M New Equilibrium 0.000 M 0.000 M 0.000 M Choose chlorous acid, HC1O2, from the list above and press the Equilibrate button. When the 0.10 M solution of HCI02 is allowed to come to equilibrium, what percent of the acid molecules are ionized? Next》 (1 of 5) Recheck 3rd attempt

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Answer 1

ans)

from above data that

According to the given data:

The percent of the acid molcules that are ionized = {[acid]_equilibrium/[acid]_initial}*100

= {[HClO₂]_consumed/[HClO₂]_initial}*100

= {([HClO₂]_initial-[HClO₂]_equilibrium)/[HClO₂]_initial}*100

= {(0.1 - 7.09*10^-2)/0.1}*100

= (2.91*10^-2/0.1)*100

= 29.1%