Chemistry
posted by George .
A 22.3 mL sample of a weak monoprotic acid, HX, requires 50.0 mL of 0.040 M NaOH to reach the equivalence point. After the addition of 30.0 mL of NaOH, the pH is 5.30. What is the Ka of HX?

mLa x Ma = mLb x Mb
Ma = (mLb x Mb)/mLa = (50.0*0.040)/22.3 = approximately 0.09M but you should do it more accurately.
So we start with 2 mmoles HX and add 30 mL x 0.04M base or add 1.2 mmoles base.
............HX + NaOH ==> NaX + H2O
initial.....2....0.........0......0
add..............1.2...............
change....1.2...1.2.....+1.2....+1.2
equil......0.8....0.........1.2....1.2
Substitute into the HendersonHasselbalch equation
pH = pKa + log (base)/(acid)
5.30 = pKa + log (1.2/0.8)
Solve for pKa and convert to Ka.
If you don't know about the HH equation, let me know and I can show you how to do it another way.