Hi, I need help answering this question. I know I need an ICE table, but I'm not sure where else to go from there.

Question

1.    The chemistry stockroom has a solution of benzoic acid, C₆H₅CO₂H, labeled “about 0.075 M.” The exact concentration needs to be determined before using the reagent in an experiment and so you are tasked with performing a titration of 10.00 mL of the acid with 0.0250 M NaOH. You find the published acid dissociation constant of benzoic acid is K_a = 6.3 x 10^-5. 

Before starting the titration, you must first choose an indicator from the list below with an endpoint that will best visually identify the equivalence point during the titration. Support your indicator choice by calculating the pH of the titration solution at the equivalence point and stating the expected color change observed for the indicator.

Indicators found in the chemistry stock room:

Crystal Violet

Thymol Blue

Bromoresol Green

Alizarin

Bromthymol Blue

Phenolphthalein

Answer 1

The first thing you should note is that you are titrating a WEAK acid (benzoic acid) with a STRONG base (NaOH). This means that the pH at equivalence will be alkaline (pH >7). So with this in mind, we can find the pH at equivalence and then choose the appropriate indicator to use. (Not really sure you need an ICE table)

Let HA represent the benzoic acid:

HA + OH^- ==> H₂O + A^-

10 ml of 0.075 M HA is equal to 0.00075 moles HA

0.00075 mols HA requires 0.00075 mols OH^-. This converts all HA to A^-. The volume OH^- required will be

0.00075 moles OH^- x 1 L / 0.025 mol = 0.03 L = 30 mls. Now total volume = 30 ml + 10 ml = 40 mls

[A^-] = 0.00075 mols / 0.040 L = 0.01875 M

Now look at the hydrolysis of A^-:

A^- + H₂O == HA + OH^- and now A^- is acting as a base. So Kb = 1x10^-14/Ka = 1x10^-14/6.3x10^-5 = 1.6x10^-10

Kb = 1.6x10^-10 = [HA][OH^-] / [A^-] = (x)(x) / 0.01875

x² = 3x10^-12

x = 1.73x10^-6 M = [OH^-]

pOH = -log 1.73x10^-6 M = 5.76

pH = 14 - 5.76

pH = 8.23

Now look up which indicator is useful at this pH