A chemist dissolves 240mg of pure barium hydroxide in enough water to make up of solution. Calculate the pH of the solution. (The temperature of the solution is .)

pH = 12.22

Explanation:

... To make up 170mL of solution... The temperature is 25°C...

The dissolution of Barium Hydroxide, Ba(OH)₂ occurs as follows:

Ba(OH)₂ ⇄ Ba²⁺(aq) + 2OH⁻(aq)

Where 1 mole of barium hydroxide produce 2 moles of hydroxide ion.

To solve this question we need to convert mass of the hydroxide to moles with its molar mass. Twice these moles are moles of hydroxide ion (Based on the chemical equation). With moles of OH⁻ and the volume we can find [OH⁻] and [H⁺] using Kw. As pH = -log[H⁺], we can solve this problem:

Moles Ba(OH)₂ molar mass: 171.34g/mol

0.240g * (1mol / 171.34g) = 1.4x10⁻³ moles * 2 =

2.80x10⁻³ moles of OH⁻

Molarity [OH⁻] and [H⁺]

2.80x10⁻³ moles of OH⁻ / 0.170L = 0.01648M

As Kw at 25°C is 1x10⁻¹⁴:

Kw = 1x10⁻¹⁴ = [OH⁻] [H⁺]

[H⁺] = Kw / [OH⁻] = 1x10⁻¹⁴/0.01648M = 6.068x10⁻¹³M

pH:

pH = -log [H⁺]

pH = -log [6.068x10⁻¹³M]

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