Answer to Question #93856 in General Chemistry for Kylie

Question

(c) Suppose we titrate 20 mL of 0.100M HCHO2 with 0.100M NaOH. Calculate the pH:
I. Before any base is added.
II. When half of the HCHO2 has been neutralized.
III. After a total of 15.0mL of base has been added.

Answer 1

You must know that when you react a weak acid with a strong base, a salt is produced.. The salt dissociates to produce the conjugate base of the acid in solution. As long as there is excess acid, you have a solution of weak acid and its conjugate base dissolved in it . This is the makings of a buffer solution.

1) First job is to calculate the [H+] of the acid solution.

Ka = [H+]² /[HCOOH]

1.8*10^-4 = [H+]² / 0.100

[H+]² = 1.8*10^-5

[H+] = 4.24*10^-3

pH = - log 4.24*10^-3

pH = 2.37 That is the first answer

Note : at this stage no buffer solution has been prepared.

Now you add sufficient NaOH to half neutralize the acid. There is a very easy way to get this pH, but because you are learning, I will do it the long way:

Because the acid and base solutions have the same molarity they react 1:1 volume basis.

To half neutralise the 20mL acid you must add 10mL of base

Mol NaOH in 10mL of 0.100M NaOH solution = 10/1000*0.1 = 0.001 mol NaOH

This will produce 0.001 mol HCOONa and leave 0.001 mol HCOOH unreacted

Your solution has a volume = 30mL

Molarity of acid = 0.001/0.030 = 0.0333M

Molarity of conjugate base = 0.001/0.030 = 0.0333M

To determine the pH yiu use the Henderson - Hasselbalch equation:

pKa HCOOH = -log ( 1.8*10^-4) = 3.74

pH = pKa + log ([conj base) / [acid] )

pH = pKa + log 0.0333/0.0333

pH = 3.74 + log 1.0

pH = 3.74 + 0

pH = 3.74

The easy way: compare the pH with the pKa of the acid: This is always true when [conj base] =[acid] which occurs when the acid is half neutralized.

The final part:

You add 15mL of base:

mol NaOH in 15.0mL of 0.1M NaOH solution = 15/1000*0.1 = 0.0015 mol NaOH

This will produce 0.0015 mol salt ( conjugate base) and 0.0005 mol acid will remain unreacted.

These amounts are dissolved in 35.0mL solution

Molarity of:

Conj base = 0.0015/0.035 = 0.0428M

Acid = 0.0005/0.035 = 0.0143M

pH = pKa + log ([conj base]/[acid])

pH = 3.74 + log (0.0428/0.0143)

pH = 3.74 + log 2.996

pH = 3.74 + 0.48

pH = 4.22