The concentration of the hydroxide ions, OH¯ in 250 mL of the solution containing the maximum amount of dissolved calcium hydroxide is 0.01728 M
We'll begin by calculating the number of mole of in 0.16 g of Ca(OH)₂. This can be obtained as follow:
Mass of Ca(OH)₂ = 0.16 g
Molar mass of Ca(OH)₂ = 40 + 2[16 + 1] = 74 g/mol
<h3>Mole of Ca(OH)₂ =? </h3>
Mole = mass / molar mass
Mole of Ca(OH)₂ = 0.16 / 74
<h3>Mole of Ca(OH)₂ = 0.00216 mole </h3>
- Next, we shall determine the molarity of the stock solution of Ca(OH)₂.
Mole of Ca(OH)₂ = 0.00216 mole
Volume = 100 mL = 100 / 1000 = 0.1 L
<h3>Molarity of Ca(OH)₂ =? </h3>
Molarity = mole / Volume
Molarity of Ca(OH)₂ = 0.00216 / 0.1
<h3>Molarity of Ca(OH)₂ = 0.0216 M</h3>
- Next, we shall determine the molarity of the diluted solution. This can be obtained as follow:
Volume of stock solution (V₁) = 100 mL
Molarity of stock solution (M₁) = 0.0216 M
Volume of diluted solution (V₂) = 250 mL
<h3>Molarity of diluted solution (M₂) =?</h3>
<h3>M₁V₁ = M₂V₂</h3>
0.0216 × 100 = M₂ × 250
2.16 = M₂ × 250
Divide both side by 250
M₂ = 2.16 / 250
<h3>M₂ = 0.00864 M</h3>
Thus, the molarity of the diluted solution is 0.00864 M
- Finally, we shall determine the concentration of the hydroxide ions, OH¯ in the diluted solution. This can be obtained as follow:
Ca(OH)₂(aq) —> Ca²⁺(aq) + 2OH¯(aq)
From the balanced equation above,
1 mole of Ca(OH)₂ contains 2 moles of OH¯
Therefore,
0.00864 M Ca(OH)₂ will contain = 2 × 0.00864 = 0.01728 M OH¯
Thus, the concentration of the hydroxide ions, OH¯ in 250 mL of the solution containing the maximum amount of dissolved calcium hydroxide is 0.01728 M
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