the ph of a 0.1molar solution of the acid HQ is 3.the value of the ionization constant ka of the acid is
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HQ will dissociate as
H2O+HQ↔H3O^+Q^-
Ka=(H3O+)(Q)/(HQ)
-log(H30+)=pH
-log(H3O+)=3
(H3O+)=10^-3
(H3O+)=(Q^-)=10^-3
Ka=(10^-3)/(0.1-10^-3)=(10^-3)²/(0.1)=1×10^-5
therefore correct answer is 1×10^-5
H2O+HQ↔H3O^+Q^-
Ka=(H3O+)(Q)/(HQ)
-log(H30+)=pH
-log(H3O+)=3
(H3O+)=10^-3
(H3O+)=(Q^-)=10^-3
Ka=(10^-3)/(0.1-10^-3)=(10^-3)²/(0.1)=1×10^-5
therefore correct answer is 1×10^-5
Answered by
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heya......
your answer
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the correct answer is 1 x 10^-5
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hope it helps ^_^
your answer
===========
the correct answer is 1 x 10^-5
=============
hope it helps ^_^
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