Question

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?

Answer 1

moles of acid HA = Molarity X volume

moles of acid=0.1 X 75 mmoles = 7.5 mmoles

Moles of base added=Molarity X volume

Moles of base added =0.1 X 30= 3 mmoles

tb

he strong base added will react with weak acid to form a salt.

ANa

the moles of salt formed is = moles of base added

= 3 mmoles

moles of acid left = 7.5-3= 4.5

this will result in formation of a buffer

the pH of buffer is given by

pH = pKa + log(salt/ acid)

given pH is 5.5

therefore

5.5= pKa + log(3/4.5)

5 .5= pKa + (-0.176)

pKa = 5.68

Answer 2

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