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)
13
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

i believe it is c.) 1575 j.

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