Question

A compound has a pKa of 7.4. To 100mL of a 1.0 M solution of this compound at pH 8.0 is added 30 mL of 1.0M hydrochloric acid. What is the resulting pH?

Answer 1

A compound has a pKa of 7.4. To 100mL of a 1.0 M solution of this compound at pH 8.0 is added 30 mL of 1.0M hydrochloric acid. What is the resulting pH?

Given : pka of an acid = 7.4

Volume = 100 mL

Concentration = 1.0 M

pH = 8.0

Volume of HCl = 30 mL

Concentration 1.0 mL

Solution:

Lets calculate ka of an acid from its pka

We know

Pka = -log ka

Ka = antilog ( -pka)

= (antilog (-7.4)

= 3.98 X 10^-8

Ka value tells us the acid is weak acid .

Lets assume given acid is HA

We use Henderson equation to calculate resulting pH

From given pH we calculate [Base]/[Acid]

pH = pka + log ([A-]/[HA])

8 = 7.4 + log [A-]/[HA]

log [A-]/[HA]= 0.6

[A-]/[HA]= Antilog 0.6

[A-]/[HA]=3.98

Or we can say that mole ratio of A- to HA is 3.98

This suggests us that [A-] = 3.94 * [HA]

Or moles of A- = 3.98 * moles of HA

Lets find moles of HA

# moles of HA = 1.0 M * 0.1 L

= 0.1 mol

Therefore

# moles of A- = 3.98 * 0.1

= 0.398 mol A-

Now we got moles of A- and HA

Lets use concentration of HCl to get concentration of H+

Concentration of HCl = [H+] (Strong acid )

[H+]= 1.0 M

Now # moles of H+ =1.0 M * 0.030 L

= 0.030 mol

Since H+ will react with A- and we will get new moles of A- , HA and also H+

•                    +               H+ --