Answer:
see below
Explanation:
[H₃O⁺] = 10^-pH
a) Soln X with pH = 4.9 => [H₃O⁺] = 10⁻⁴°⁹ = 2.51 x 10⁻⁵M
b) Soln Y with pH = 7.4 => [H₃O⁺] = 10⁻⁷°⁴ = 3.98 x 10⁻⁸M
Answer:
a. [H30+] = 4 * 10^-10 M
b. [H3O+]= 4 * 10^-8 M
Explanation:
Step 1: Data given
pH of solution X = 9.4
pH of solution Y = 7.4
Step 2: Calculate [H3O+] of solution X
pH = -log [H+]
[H+] = [H30+]
pH = -log[H3O+]
9.4 = -log[H3O+]
10^-9.4 = 3.98*10*-10
[H30+] = 4 * 10^-10 M
To control we can calculate pH
-log[4 * 10^-10] = 9.4
Step 3: Calculate [H3O+] of solution Y
pH = -log [H+]
[H+] = [H30+]
pH = -log[H3O+]
7.4 = -log[H3O+]
10^-7.4 = 4*10*-8
[H30+] = 4 * 10^-8 M
To control we can calculate pH
-log[4 * 10^-8] = 74
a. [H30+] = 4 * 10^-10 M
b. [H3O+]= 4 * 10^-8 M
