pKa = -log Ka.
HA + H2O ==> H3o^+ + A^-
Ka = (H3O^+)(A^-)/(HA)
Then (A^-)/(HA) = Ka/(H3O^+).
Therefore, if pH = 5, then (H3O^+) = 1E-5. Substitute into Ka expression and (A^-)/(HA) = 6.3E-3.
When that ratio is = or > 10 you get the A^- color; when the ratio is = or < 1/10 you get the HA color.
A certain indicator, HA, has a Ka value of 6.3 × 10-8. The protonated form of the indicator is blue and the ionized form is red.
-What is the PKa of the indicator?
-What is the color of the indicator in a solution with pH=5?
1 answer
1 answer