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Answer:
A. Ba(OH)₂ (aq) <==> Ba²⁺ (aq) + 2OH¯ (aq)
B. The molar concentration of OH¯ is 0.072 M
C. The pH of the solution is 12.86
Explanation:
A. Dissociation equation
Barium hydroxide Ba(OH)₂ dissolves in water by the following equation:
Ba(OH)₂ (aq) <==> Ba²⁺ (aq) + 2OH¯ (aq)
B. Determination of the molar concentration of OH¯.
We'll begin by calculating the number of mole in 1.5 g of Ba(OH)₂. This can be obtained as follow:
Mass of Ba(OH)₂ = 1.5 g
Molar mass of Ba(OH)₂ = 137 + 2(16 + 1)
= 137 + 2(17)
= 137 + 34
= 171 g/mol
Mole of Ba(OH)₂ =?
Mole = mass / Molar mass
Mole of Ba(OH)₂ = 1.5 / 171
Mole of Ba(OH)₂ = 0.009 mole
Next, we shall determine the molarity of Ba(OH)₂. This can be obtained as follow:
Mole of Ba(OH)₂ = 0.009 mole
Volume of solution = 0.250 L
Molarity of Ba(OH)₂ =?
Molarity = mole / Volume
Molarity of Ba(OH)₂ = 0.009 / 0.250
Molarity of Ba(OH)₂ = 0.036 M
Finally, we shall determine the molar concentration of OH¯.this can be obtained as follow:
Ba(OH)₂ (aq) <==> Ba²⁺ (aq) + 2OH¯ (aq)
From the balanced equation above,
1 mole of Ba(OH)₂ produced 2 moles OH¯.
Therefore, 0.036 M Ba(OH)₂ will produce = 0.036 × 2 = 0.072 M OH¯.
Therefore, the molar concentration of OH¯ is 0.072 M
C. Determination of the pH.
We'll begin by calculating the pOH of the solution. This can be obtained as follow:
Concentration of OH¯, [OH¯] = 0.072 M
pOH =?
pOH = –Log [OH¯]
pOH = –Log 0.072
pOH = 1.14
Finally, we shall determine the pH. This can be obtained as follow:
pOH = 1.14
pH =?
pH + pOH = 14
pH + 1.14 = 14
Collect like terms
pH = 14 – 1.14
pH = 12.86