Question

Calculate the pH of each of the following strong acid solutions.

1. 20.00 mL of 1.50 M  HCl diluted to 0.480 L . Express the pH of the solution to three decimal places.

2. A mixture formed by adding 46.0 mL of 2.5×10⁻²M  HCl to 120 mL of 1.5×10⁻²M  HI. Express the pH of the solution to two decimal places.

Answer 1

1)

1st find the concentration of diluted acid

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M1 = 1.5 M

V1 = 20.0 mL

V2 = 0.480 L = 480 mL

use:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (1.5*20)/480

M2 = 0.0625 M

use:

pH = -log [H+]

= -log (6.25*10^-2)

= 1.2041

Answer: 1.204

2)

Concentration of mixture = (n1*C1*V1+ n2*C2*V2) / (V1+V2)

where C1 --> Concentration of 1 component

V1-->volume of 1 component

C2 --> Concentration of other component

V2-->volume of other component

n1 --> number of particle from 1 molecule of 1st component

n2 --> number of particle from 1 molecule of 2nd component

use:

[H+] = (n1*C1*V1+ n2*C2*V2) / (V1+V2)

= (1*0.025*46+1*0.015*120)/(46+120)

= 0.0178 M

use:

pH = -log [H+]

= -log (1.78*10^-2)

= 1.7496

Answer: 1.75