Part 1: A chemist performs an acid–base titration using 0.188 M aqueous NaOH as

Question

Part 1:

A chemist performs an acid–base titration using 0.188 M aqueous NaOH as the titrant with 25.75 mL of 0.102 M H2SO4. The endpoint of the titration is detected using phenolphthalein as the indicator. The balanced equation for the reaction is shown below:

2NaOH(aq)+H2SO4(aq) 2NaCl(aq)+2H2O(l)

What volume of NaOH is expected to be required to consume all of the acid initially present?

Part 2:

alizarine yellow; pKa = 11.0

phenolpthalein; pKa = 9.2

bromothymol blue; pKa = 6.8

phenol red; pKa = 7.2

methyl yellow; pKa = 3.5

Explanation / Answer

Part 1

2NaOH(aq)+H2SO4(aq) 2NaCl(aq)+2H2O(l)

Given that

NaOH = 0.188 M

Voluem of NaOH = ?

H2SO4 = 0.102 M

Volume = 25.75

And

2NaOH(aq)+H2SO4(aq) 2NaCl(aq)+2H2O(l)

First calculate the mole of H2SO4 as follows:

Number of moles = molarity * volume in L

= 0.102 * 25.75/1000

= 0.00263 Mole H2SO4

Then calculate the mole of NaOH which are reacted with H2SO4:

0.00263 Mole H2SO4 *2/1

= 0.005253 Mole NaOH

We know that ; molarity = number of moles / volume in L

Volume in L = 0.005253 Mole NaOH /0.188 M

= 0.0279 L

= 27.9 ml

Hence 27.9 ml NaOH is expected to be required to consume all of the acid initially present.

Part 2:

bromothymol blue and phenol red

Indicator:

Indicator of acid-base titration is a compound that changes color according to pH changes of solution. In general the color of indicator of acid-base titration changes when the acidic solution becomes basic or basic solution becomes acidic.

The following indicators which are use to determine the pH of the solution more precisely:

Methyl violet

Thymol blue

Methyl orange

Methyl red

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