Question

What mass of LiOH would need to be dissolved in water to make 300.0 mL of a solution with a pH of 11.33?

Answer 1

Execute Order 66________________

Answer 2

The mass of LiOH needed to make a 300 mL of a solution with a pH of 11.33 is 0.015 g

We'll begin by calculating the pOH of the solution.

pH = 11.33

pOH =?

pH + pOH = 14

11.33 + pOH = 14

Collect like terms

pOH = 14 – 11.33

pOH = 2.67

• Next, we shall determine the concentration of hydroxide ion, OH⁻

pOH = 2.67

Concentration of hydroxide ion [OH⁻] = ?

pOH = –Log [OH⁻]

2.67 = –Log [OH⁻]

–2.67 = Log [OH⁻]

Take the as antilog of –2.67

[OH⁻] = antilog (–2.67)

[OH⁻] = 0.0021 M

• Next, we shall determine the concentration of LiOH.

LiOH(aq) —> Li⁺(aq) + OH⁻(aq)

From the balanced equation above,

1 mole of LiOH contains 1 mole of OH⁻

Therefore,

0.0021 M LiOH will also contain 0.0021 M OH⁻.

• Next, we shall determine the mole of LiOH in the solution.

Molarity of LiOH = 0.0021 M

Volume = 300 mL = 300 / 1000 = 0.3 L

Mole of LiOH =?

Mole = Molarity × volume

Mole of LiOH = 0.0021 × 0.3

Mole of LiOH = 0.00063 mole

• Finally, we shall determine the mass of LiOH needed to prepare the solution.

Mole of LiOH = 0.00063 mole

Molar mass of LiOH = 7 + 16 + 1 = 24 g/mol

Mass of LiOH = ?

Mass = mole × molar mass

Mass of LiOH = 0.00063 × 24

Mass of LiOH = 0.015 g

Therefore, the mass of LiOH needed to prepare the solution is 0.015 g

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