Question

Phenolphhalein indicator is used in this experiment because it changes color in the pH range of 8.0 to 9.6.

Show by calculation that phenolphthalein is an acceptable indicator for the titration of KHP with NaOH.

This is true if the pH at the equivalence point falls within this range.

In essence, what you need to do is to calculate the pH of the unprotonated phthalate salt. You may assume that the concentration of the salt is approximately 0.05 M at the equivalance point.

HINT: The salt is derived from the weak diprotic acid, which has Ka1 = 1.2 * 10^-3 and Ka2 = 3.9 * 10^-6

Answer 1

assume salt is Na2P

P-2 + H2O ----------------------> HP-   + OH-

0.05                                            0           0 ------------------> initial

0.05-x                                        x           x --------------------> equilibrium

Kb1 = [HP-][OH-]/[P-2]

Kw / Ka2 = x^2 / 0.05-x

1.0 x 10^-14 / 3.9 x 10^-6 = x^2 / 0.05-x

2.56 x 10^-9 = x^2 / 0.05-x

x^2 + 2.56 x 10^-9 x - 1.28 x 10^-10 = 0

x = 1.13 x 10^-5

[OH-] = x = 1.13 x 10^-5 M

pOH = -log [OH-]

pOH = -log (1.13 x 10^-5)

pOH = 4.95

pH + pOH = 14

pH = 9.05