1)what is the ph of .12M solution of C4H6O2 (ka= 1.45 x10^-5)... this is how i d
ID: 75736 • Letter: 1
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1)what is the ph of .12M solution of C4H6O2 (ka= 1.45 x10^-5)... this is how i did it:: C4H6O2 + H2o <-->H3o+ C4h5O2- (X2/.12-X) = 1.45 x 10^-5 X= .0013 PH= -log .0013= 2.88 2) what is PH of a .7 M solution of C4H11N (Kb= 5.89 x10^-4) C4H11N + H2o <--> C4H10N+ + OH- X2/.7-X = 5.89 x 10^-4 X= .0203 Poh= -log .0203= 1.69... Ph= 14-1.69 = 12.3 PLEASE TELL ME IF THIS IS CORRECT! 1)what is the ph of .12M solution of C4H6O2 (ka= 1.45 x10^-5)... this is how i did it:: C4H6O2 + H2o <-->H3o+ C4h5O2- (X2/.12-X) = 1.45 x 10^-5 X= .0013 PH= -log .0013= 2.88 2) what is PH of a .7 M solution of C4H11N (Kb= 5.89 x10^-4) C4H11N + H2o <--> C4H10N+ + OH- X2/.7-X = 5.89 x 10^-4 X= .0203 Poh= -log .0203= 1.69... Ph= 14-1.69 = 12.3 PLEASE TELL ME IF THIS IS CORRECT!Explanation / Answer
1)what is the ph of .12M solution of C4H6O2 (ka= 1.45 x10^-5)... this is how i did it:: C4H6O2 + H2o <-->H3o+ C4h5O2- (X2/.12-X) = 1.45 x 10^-5 X= .0013 this is the correctvalue. If you plug it back into the above expression, you get:( 0.0013190906^2)/(0.12- 0.0013190906) = 1.47 x10^-5
PH= -log .0013= 2.88 Yes, takethe negative log of x = [H3O+]
2) what is PH of a .7 M solution of C4H11N (Kb= 5.89 x10^-4) C4H11N + H2o <--> C4H10N+ + OH- X2/.7-X = 5.89 x 10^-4 X= .0203
(0.0203)^2/(0.7-0.0203) = 6.1e-4, so the answeris quite close
Poh= -log .0203= 1.69... Ph= 14-1.69 = 12.3 Yes, theseare the correct equations.
PLEASE TELL ME IF THIS IS CORRECT!
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