You have 75.0 ml of 0.10 m ha. after adding 30.0 ml of 0.10 m naoh, the ph is 5.50. what is the ka value of ha?

First, we need to get moles of HA =molarity * volume 
                                                        = 0.1 m * 0.075 L = 0.0075 moles

moles of NaOH = molarity * volume
                           = 0.1 * 0.03 L  = 0.003 moles 

from the reaction equation:

HA(aq) + NaOH(aq)  → NaA(aq)  + H2O(l)

that means the final moles' HA = 0.0075 - 0.003 =0.0045 moles

when the total volume is = 0.075 + 0.03 L = 0.105 L

∴ [HA] = moles / volume

           = 0.0045 / 0.105 L = 0.043 m

[A^-] = 0.003 / 0.105 L = 0.029 m

then by using H-H equation:

PH = Pka + ㏒[A^-] / [HA]

by substitution, we can get Pka:

5.5 = Pka+ ㏒ (0.029 /0.043)

∴ Pka = 5.67 

when Pka = - ㏒Ka

         5.67 = -㏒ Ka 

∴Ka = 2 x 10^-6