An aqueous solution is produced by dissolving 1.5 g of barium hydroxide (Ba(OH)2(aq) in enough water to make 0.250 L.
a) write the dissociation reaction of barium hydroxide
b) calculate the molar concentration of OH(aq)
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c) calculate pH

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