Question

Ignore activity effects for the following questions: A buffer is prepared by adding 25.00 mL of...

Ignore activity effects for the following questions:

A buffer is prepared by adding 25.00 mL of 0.100 M NaOH to 0.2525 g of benzoic acid, and diluting to 100.0 mL. What is the pH?

Homework Answers

Answer #1

Mass of benzoic acid =  0.2525 g

Mass of NaOH = 0.100 M

We know molar mass of benzoic acid = 122.12 g / mol

The moles of benzoic acid = 0.2525 / 122.12

= 0.00206

moles of NaOH = 0.100 x 25 / 1000

= 0.0025

after combining both we get ,

moles of acid = 0.00206 - 0.0025

= - 0.02294

moles of salt = 0.0025

[acid] = - 0.02294 / 0.1

= -0.2294M

[salt] = 0.0025 / 0.1

= 0.025 M

We know , pKa of benzoic acid = 4.20

Let us consider ,

pH = pKa + log [salt] / [acid]

pH = 4.20 + log [0.025] / [-0.2294]

pH = 5.16

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