Answer:
0.0140 M H₂C₆H₆O₆.
Explanation:
- We should mention the relation: <em>Ka = α²C,</em>
Where, Ka is the dissociation constant of the acid.
α is the degree of ionization of the acid.
C is the concentration if the acid.
<em>The percent of ionization (α %) = α x 100.</em>
α = √Ka/C
∴ α is inversely proportional to the concentration of the acid.
<em>So, the acid with the lowest concentration has the greatest percent ionization.</em>
Answer:
pH = 12.80
[H3O+] = 1.58 * 10^-13 M
[OH-] = 0.063 M
Explanation:
Step 1: Data given
pOH = 1.20
Temperature = 25.0 °C
Step 2: Calulate pH
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 1.20 = 12.80
Step 3: Calculate hydronium ion concentration
pH = -log[H+] = -log[H3O+]
12.80 = -log[H3O+]
10^-12.80 = [H3O+] = 1.58 * 10^-13 M
Step 4: Calculate the hydroxide ion concentration
pOH = 1.20 = -log [OH-]
10^-1.20 = [OH-] = 0.063M
Step 5: Control [H3O+] and [OH-]
[H3O+]*[OH-] = 1* 10^-14
1.58 *10^-13 * 0.063 = 1* 10^-14