........HA ==> H^+ + A^-
I....0.001.....0.....0
C......-x......x.....x
E....0.001-x...x.....x
Substitute the E line into the Ka expression and solve for x = (H^+) and convert to pH. Be careful, you may need to use the quadratic equation; i.e., you may not be able to call 0.001-x just 0.001.
pKa = -log Ka.
the ka value of 0.001moldm-3 solution of a weak acid is 1.6*10-6 calculate the ph and the pka
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