This is a buffer solution.
(A^-) is 0.35 M in the solution. (HA) is 0.81 M
.................HA ==> H^+ + A^-
I.................0.81......0.........0.35
C................-x...........x..........x
E..............0.81-x.......x........0.35+x
Ka = (H^+)(A^-)/(HA)
Plug the E line nto the Ka expression and solve for (H^+), then convert it to pH.
You may also use the Henderson-Hasselbalch equation of
pH = pKa + log (base)/(acid) where base is the A^- and the acid is HA
What is the pH of a solution containing 0.81M HA and 0.35M A- ? Ka HA=5.45x10-8.
1 answer
1 answer