Question

A 0.5-L unbuffered solution needs the pH adjusted from 6 to 5. How many microliters of a 2 molar HCl solution need to be added to adjust the pH? Report your answer to four decimal places.

Answer 1

Solution :

> Initial pH = 6 and volume = 0.5 L

Thus,

[H+] = 10^-pH = 10^-6 M

> Final pH = 5

Hence, [H+] = 10^-pH = 10^-5 M

According to molarity equation,

Molarity of mixture (M3) = M1V1 + M2V2 / V1 + V2

Given and calculated values are:

M3 = 10^-5 M

M1 = 10^-6 M

V1 = 0.50 L = 500 mL

M2 = 2.0 M

V2 = ?

Therefore,

10^-5 = (10^-6 x 500 mL + 2.0 M x V2) / (500 + V2)

5 x 10^-3 + 10^-5 V2 = 5 x 10^-4 + 2 V2

2 V2 - 0.00001 V2 = 5 x 10^-3 - 5.0 x 10^-4

1.99999 V2 = 50 x 10^-4 - 5.0 x 10^-4

1.99999 V2 = 45 x 10^-4

1.99999 V2 = 0.0045

V2 = 0.00225 mL

V2 = 0.0023 mL