How much hbro must be added to 1l of pure water to make a solution with a ph of 4.25? ka = 2.00 * 10-9?

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Inchworm

Chemistry

10 May, 10:48

How much hbro must be added to 1l of pure water to make a solution with a ph of 4.25? ka = 2.00 * 10-9?

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Derick Short · Answer 1

Answer is: 153.52 grams of hypobromous acid must be added.

Chemical dissociation: HBrO ⇄ H⁺ + BrO⁻.

pH = 4.25.

pH = - log[H⁺].

[H⁺] = 10∧ (-pH).

[H⁺] = 10∧ (-4.25).

[H⁺] = [BrO⁻] = 5.62·10⁻⁵ M.

Ka = [H⁺] · [BrO⁻] / [HBrO].

2.00·10⁻⁹ = (5.62·10⁻⁵ M) ² / [HBrO].

[HBrO] = 3.16·10⁻⁹ M² / 2.00·10⁻⁹.

[HBrO] = 1.58 M.

m (HBrO) = n (HBrO) · M (HBrO).

m (HBrO) = 1.58 mol · 96.91 g/mol.

m (HBrO) = 153.52 g.