pKa = -log(Ka) = 6.6
Ka = about 2.6E-7 but you need to all of
these calculations more accurately.
.............HA ==> H^+ + A^-
I............100....0......0
C...........39.6...39.6...39.6
E..........60.4....39.6...39.6
Ka = (H^+)A^-)/(HA)
Substitute the E line into the Ka expression and solve for H^+, then convert to pH.
What is the pH of a solution of a monoprotic acid with a pKa of 6.6 if it is 39.3 % ionized in solution? Just provide the numerical answer, do not prefix with pH. Give your answer to two decimal places. Remember the definition of Ka
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