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22/09/2021
Based on these values the pka = 0.967
Like many chemistry problems, the first thing to do is to write a chemical formula:
##HClO_2+H_2O ->ClO_2^(-) + H_3O^+##
Then we can write an ICE table, which just means that we write down for each non-solvent species (water in this case) the molarities of the initial state, the change in moles needed to get to equilibrium, and the resulting equilibrium molarities.
##HClO_2+H_2O ->ClO_2^(-) + H_3O^+##
I. 0.1…………………………..0…………….0
C. -x………………………….+x…………..+x
E. 0.1 – x……………………x………………x
So this table (I don’t know how to insert a good looking table in here) says that we start out with 0.1 M chlorous acid and virtually no other relevant species (we don’t analyze water for acid/base equilibrium calculations). Based on stoichiometry, we know that some chlorous acid will go away (-x), and the same number of moles will be generated of H30+ and the conjugate base (chlorite ions). This means that at equilibrium there will be the concentrations in the bottom row (sum of the I. and C. rows).
We also know the pH.
##pH=-log[H_3O^+]_(eq)##
Rearranging to calculate the hydronium ion concentration
##[H_3O^+]_(eq)=10^(-pH)##
We already found out that ##[H_3O^+]_(eq)=x## from the bottom row of the ICE table. So we already know our value for x in the above ICE table. In this case x = 0.0631 M (calculated from the pH expression).
Then we just use the expression for ##K_a## and ##pK_a## .
##K_a=([H_3O^+]_(eq)[ClO_2^-]_(eq))/[HClO_2]_(eq)=x^2/(0.1-x)=0.1079##
##pK_a=-log(K_a)=0.967##
Google lists pka = 1.94 for chlorous acid. So either the problem uses data that’s off or there is some kind of non-ideal way that this acid behaves in reality.
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I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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