Answer:
1. pOH = 1.01 and the pH = 12.99.
2. a) %I = 0.417 %, b) pH = 2.34.
Explanation:
1. The pH of the dissolution of Ba(OH)₂ in water can be calculated using the following equation:
[tex]pH = 14 - pOH = 14 - (-log[OH^{-}]) = 14 + log[OH^{-}][/tex]
To find the pOH, first, we need to calculate the concentration (C) of Ba(OH)₂:
[tex] C_{Ba(OH)_{2}} = \frac{\eta}{V} = \frac{m}{M*V} [/tex]
Where:
η: is the number of moles of Ba(OH)₂
V: is the volume of the solution = 100 ml = 0.100 L
M: is the molar mass of Ba(OH)₂ = 171.34 g/mol
m: is the mass of Ba(OH)₂ = 0.837 g
[tex] C_{Ba(OH)_{2}} = \frac{m}{M*V} = \frac{0.837 g}{171.34 g/mol*0.100 L} = 0.049 mol/L [/tex]
Knowing that in 1 mol of Ba(OH)₂ we have two moles of OH⁻, the concentration of OH⁻ is:
[tex] C_{OH^{-}} = \frac{2*\eta_{Ba(OH)_{2}}}{V} = 2*C_{Ba(OH)_{2}} = 2*0.049 mol/L = 0.098 mol/L [/tex]
Now, we can find the pOH and the pH of the solution:
[tex] pOH = -log[OH^{-}] = -log(0.098) = 1.01 [/tex]
[tex] pH = 14 - pOH = 14 - 1.01 = 12.99 [/tex]
2. a) The percentage of ionization (% I) of acetic acid can be calculated using the following equation:
[tex]\% I = \frac{[H^{+}]_{eq}}{[HA]_{0}}*100[/tex]
We need to find the [H⁺] = [H₃O⁺] at the equilibrium:
CH₃COOH(aq) + H₂O(l) → CH₃COO⁻(aq) + H₃O⁺(aq)
1.0 - x x x
[tex] Ka = \frac{[CH_{3}COO^{-}]*[H_{3}O^{+}]}{[CH_{3}COOH]} [/tex]
Ka: is the dissociation constant of acetic acid = 1.75x10⁻⁵
[tex] 1.75\cdot 10^{-5} = \frac{x^{2}}{1.0 - x} [/tex]
[tex] 1.75\cdot 10^{-5}(1.0 - x) - x^{2} = 0 [/tex] (1)
By solving equation (1) for x we have two solutions:
x₁ = -0.00417
x₂ = 0.00417 = [H₃O⁺] = [CH₃COO⁻]
We will take the positive value to find the percentage ionization:
[tex]\% I = \frac{[H^{+}]_{eq}}{[HA]_{0}}*100 = \frac{0.00417 M}{1.0 M}*100 = 0.417 \%[/tex]
b) The pH of the solution is:
[tex] pH = -log([H_{3}O^{+}]) = -log(0.00417) = 2.34 [/tex]
I hope it helps you!
