Asked by mary
Calculate the pH of 0.15M hydrazoic acid, HN3
. Ka = 1.8 x 10-5
.
. Ka = 1.8 x 10-5
.
Answers
Answered by
DrBob222
...............HN3 ==> H^+ + N^3-
I..............0.15..........0.........0
C...............-x............x..........x
E..............0.15-x......x...........x
Plug the E line (equilibrium line) into the Ka expression and solve for x = (H^+).
Then convert H^+ to pH by pH = - log (H^+). Post your work if you get stuck.
I..............0.15..........0.........0
C...............-x............x..........x
E..............0.15-x......x...........x
Plug the E line (equilibrium line) into the Ka expression and solve for x = (H^+).
Then convert H^+ to pH by pH = - log (H^+). Post your work if you get stuck.
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