Answered

From tech troubles to travel tips, IDNLearn.com has answers to all your questions. Discover trustworthy solutions to your questions quickly and accurately with help from our dedicated community of experts.

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

Sagot :

Batolisisidn

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

We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com is your source for precise answers. Thank you for visiting, and we look forward to helping you again soon.