Question

. How many grams of HOBr (solute) must be added to 1.0 L of water to produce pH=9.00? What fraction of the hyprobromous acid solute dissociates at pH=9.00?
pKa=8.60

Answer 1

Given data,

pH = 9.00 and pKa of HOBr = 8.60

Let us find out [H⁺] = ? and diissociation constant Ka= ?

We have,

pH = -log[H⁺] ........(logarithmic form)

[H⁺] = 10^-pH ....... (Index form)

[H⁺] = 10^-9

[H⁺] = 1 x 10^-9.

And,

pKa = -log(Ka)

$\Rightarrow$ Ka = 10-pKa

Ka = 10-8.60

Ka = 2.51 x 10^-9.

Pertinent dissociation equation for HOBr is,

BrOH (aq.) <----------> BrO^- (aq.) + H⁺ (aq.)

Hence corresponding expression for dissociation constatnt is,

Ka = [BrO^-][H⁺] / [BrOH] ..........(1) [H+] << [HOBr] hence [BrOH] - [H+] $\approx$ [BrOH] at denominator

It's clear from dissociation equation that,

[BrO^-] = [H⁺] = 1 x 10^-9,

With all known values in eq. (1)

2.51 x 10^-9 = (1 x 10^-9)(1 x 10^-9) / [HOBr]

[HOBr] = (1 x 10^-18) / (2.51 x 10^-9)

[HOBr] = (1 x 10^-9 /2.51

[HOBr] = 3.98 x 10^-10 M/L

Molar mass of HOBr = 16.9 g/mol

Hence

Mass of 3.98 x 10^-10 mole BrOH = 16.9 x 3.98 x 10^-10 = 6.7 x 10^-10 g.

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