* Calculate the value of ki when 3.4 moles of HX is dissolved in I liter of solution it is found at equilibrium that [H₂0]. Calculate ki for HX HX + H₂O = Hz 8 + + X -005 3.4 moles solution it is of HX when of solution found ( = 2×10²-10) calculate ki for HX [H₂0+] [04] = 1x 10-14 that % ion = 2.3% %⁰ ionization = [H₂8+] [100] initial [H 30+][X] ki= Тиху is dissoled in Lliter at equilibirun that [04]
* Calculate the value of ki when 3.4 moles of HX is dissolved in I liter of solution it is found at equilibrium that [H₂0]. Calculate ki for HX HX + H₂O = Hz 8 + + X -005 3.4 moles solution it is of HX when of solution found ( = 2×10²-10) calculate ki for HX [H₂0+] [04] = 1x 10-14 that % ion = 2.3% %⁰ ionization = [H₂8+] [100] initial [H 30+][X] ki= Тиху is dissoled in Lliter at equilibirun that [04]
Chapter16: Applications Of Neutralization Titrations
Section: Chapter Questions
Problem 16.28QAP
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![* Calculate the value of ki
I liter of
when 3.4 moles of HX is dissolved in
solution it is found at equilibrium that [H3O+]=.005
Calculate ki for HX
o
O
HX +H 2O = Hz8t +X
-005
3.4 moles
When
of solution it is
= 2×10=10 calculate ki
[1₂0+] [04] = 1x 10-14
that % ion = 2.3%
% ionization = [H₂8+] [100]
initial [HXJ
of
[H 30 + ][X=]
found
ki for HX
ki= [HX]
dissolved in
is
Lliter
at equilibirum that [OH])](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffd4c8a78-94ed-40f4-97b5-ac0eacd4f1aa%2F23536786-fe3c-45e1-9677-b6399a0d2c10%2Fqlc31g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:* Calculate the value of ki
I liter of
when 3.4 moles of HX is dissolved in
solution it is found at equilibrium that [H3O+]=.005
Calculate ki for HX
o
O
HX +H 2O = Hz8t +X
-005
3.4 moles
When
of solution it is
= 2×10=10 calculate ki
[1₂0+] [04] = 1x 10-14
that % ion = 2.3%
% ionization = [H₂8+] [100]
initial [HXJ
of
[H 30 + ][X=]
found
ki for HX
ki= [HX]
dissolved in
is
Lliter
at equilibirum that [OH])
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