Answer:
Hello your question is incomplete the pH of the solution is unknown hence I will take the pH of the solution be = 4.7
answer : Pka = 5 i.e. > 4.7
Explanation:
Student titrates :
Unknown weak acid ( HA ) with 25 mL of 0.100 M of NaOH ( to a pale pink phenolphthalein endpoint ) then titrates with 11.0 mL of 0.100 M HCL to the solution that contains neutralized unknown acid
Assuming pH of solution = 4.7
HA + NaOH --------> H2O + NaA
at equilibrium point
0.1 * 25 * 1 = n * 1 ∴ [tex]n_{ha}[/tex] = 2.5 mmoles ------ 1
also
[tex]A^-[/tex] + HCL ---------> HA + C[tex]L^-[/tex]
at equilibrium point
[tex]n_{a}[/tex] = 1.1 mmoles --------- 2
The ratio of reaction 2 to 1
[tex]\frac{n_{a} }{n_{ha} }[/tex] = 1.1 / 2.5 ≈ 1/2 ( this show that when the pH = 4.7 approximately half of the A^- have been converted to HA )
Determine the pKa value of the solution
Given that
pH = Pka + log ( A^- / hA )
4.7 = Pka + log ( 1/2 )
therefore Pka = 4.7 - ( -0.3 ) = 5
