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