Calculate the pH during the titration of 20.00 mL of 0.1000 M C6H5OH(aq) with 0.1000 M KOH(aq) after 12 mL of the base have been added.

Question

Calculate the pH during the titration of 20.00 mL of 0.1000 M C₆H₅OH(aq) with 0.1000 M KOH(aq) after 12 mL of the base have been added. K_a of phenol acid = 1.0 x 10^-10.

The answer is: 10.18

no matter how I work this problem, I still don't get the right answer.

Please show steps to the problem so I can follow along. Thank you!

Answer 1

Hi Janel,

A buffer solution contains both acid and its conjugate base in a beaker. This is a buffer solution that is prepared by partial neutralization of an acid.

Initially, the beaker contains only acid, C₆H₅OH. Then, you add a lesser amount of base which will convert some of the acid to the conjugate base but still leave some acid unconverted.

The initial beaker contains 0.0020 moles of the acid C₆H₅OH.

Then, you add 0.0012 moles of hydroxide OH^-, which removes 0.0012 moles of acid and converts it to produce 0.0012 moles of conjugate base, C₆H₅O^-. But don't forget that you still have 0.0008 moles of acid that was unreacted and remains in the beaker.

Now, use the Henderson-Hasselbalch equation:

pH = pK_a + log ( [conjugate base] / [acid] )

pH = -log(1x10^-10) + log ( [0.0012] / [0.0008] )

pH = 10.00 + 0.18

pH = 10.18

Answer 2

No. of moles [C6H5OH] = molarity * volume in L = 0.1 * .02 = .002

No. of moles [KOH] = 0.1 * 0.012 moles = .0012 mol

C6H5OH + KOH C6H5OK + H2O

I           .002

C          -.0012

E          .0008

Pka = -logKa = -log10^-10 = 10

PH = Pka + log(C6H5OK/C6H5OH)

PH= 10 + Log(.0012/.0008)

PH = 10 + 0.176 = 10.176 = 10.18