Question

1. A solution was made by dissolving 0.837g of Ba(OH)2 in 100 mL of solution. What is the pOH and pH of the solution? 


2. What is the percentage ionization of 1.0M acetic acid? What is the pH of this solution?

Answer 1

1. pOH = 1.01 and the pH = 12.99.

2. a) %I = 0.417 %, b) pH = 2.34.

Step-by-step explanation:

1. The pH of the dissolution of Ba(OH)₂ in water can be calculated using the following equation:

`pH = 14 - pOH = 14 - (-log[OH^(-)]) = 14 + log[OH^(-)]`

To find the pOH, first, we need to calculate the concentration (C) of Ba(OH)₂:

`C_{Ba(OH)_(2)} = (\eta)/(V) = (m)/(M*V)`

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

`C_{Ba(OH)_(2)} = (m)/(M*V) = (0.837 g)/(171.34 g/mol*0.100 L) = 0.049 mol/L`

Knowing that in 1 mol of Ba(OH)₂ we have two moles of OH⁻, the concentration of OH⁻ is:

`C_(OH^(-)) = \frac{2*\eta_{Ba(OH)_(2)}}{V} = 2*C_{Ba(OH)_(2)} = 2*0.049 mol/L = 0.098 mol/L`

Now, we can find the pOH and the pH of the solution:

`pOH = -log[OH^(-)] = -log(0.098) = 1.01`

`pH = 14 - pOH = 14 - 1.01 = 12.99`

2. a) The percentage of ionization (% I) of acetic acid can be calculated using the following equation:

`\% I = ([H^(+)]_(eq))/([HA]_(0))*100`

We need to find the [H⁺] = [H₃O⁺] at the equilibrium:

CH₃COOH(aq) + H₂O(l) → CH₃COO⁻(aq) + H₃O⁺(aq)

1.0 - x x x

`Ka = ([CH_(3)COO^(-)]*[H_(3)O^(+)])/([CH_(3)COOH])`

Ka: is the dissociation constant of acetic acid = 1.75x10⁻⁵

`1.75\cdot 10^(-5) = (x^(2))/(1.0 - x)`

`1.75\cdot 10^(-5)(1.0 - x) - x^(2) = 0` (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:

`\% I = ([H^(+)]_(eq))/([HA]_(0))*100 = (0.00417 M)/(1.0 M)*100 = 0.417 \%`

b) The pH of the solution is:

`pH = -log([H_(3)O^(+)]) = -log(0.00417) = 2.34`

I hope it helps you!