pH = 5.50
pH = -log(H^+)
(H^+) = approx 3.2E-6 but you need a more accurate answer than that.
.........HA ==> H^+ + A^-
I.........y.....0......0
C........-x.....x......x
E........y-x....x......x
Ka = (H^+)(A^-)/(HA)
5.7E-10 = (x)(x)/(y-x)
You know x, substitute and solve for y.
What is the concentration of a monoprotic weak Acid if its pH is 5.50 and its Ka= 5.7 x 10^-10?
1 answer
1 answer