Question

calculate the pH of the following

a) 20.00 mL of 1.40 M HCl diluted to .500 L

b) a mixture formed by adding 44.0 mL of 2.5 × 10^-2 M HCl to 150 mL of 1.0 ×10^-2 M HI

Answer 1

a)
use dilution formula
M1*V1 = M2*V2
1---> is for stock solution
2---> is for diluted solution

Given:
M1 = 1.4 M
V1 = 20 mL
V2 = 500 mL

use:
M1*V1 = M2*V2
M2 = (M1*V1)/V2
M2 = (1.4*20)/500
M2 = 0.056 M

Given:
[H+] = 5.6*10^-2 M

use:
pH = -log [H+]
= -log (5.6*10^-2)
= 1.2518
Answer: 1.25

b)
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

Here:
n1 = 1
n2 = 1

use:
C = (n1*C1*V1+ n2*C2*V2) / (V1+V2)
C = (1*0.025*44+1*0.01*150)/(44+150)
C = 0.0134 M

Given:
[H+] = 1.34*10^-2 M

use:
pH = -log [H+]
= -log (1.34*10^-2)
= 1.8729
Answer: 1.87