A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein endpoint with 25.0 mL of 0.100 M NaOH. The student then titrates in 11.0 mL of 0.100 M HCl to the solution containing the neutralized unknown acid (HCl was delivered from a second buret). The pH of the resulting solution is pH

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  ∴   = 2.5 mmoles ------ 1

also

 + HCL ---------> HA + C

at equilibrium point

= 1.1 mmoles --------- 2

The ratio of reaction 2 to 1

= 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

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

the answer to your question is c