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Suani0822
reaction is NH4OH <-> NH4+ OH- (note this is reversible)

Draw up an ICE table

Let x be equilibrium conc of OH- assume init conc of OH is 0M and init conc of NH4+ is 0M also. Init conc of NH4OH is 0.1M so equilibrium conc will be 0.1-x.

%dissociation = x/0.1-x * 100%

1 = 100x/0.1-x

0.1-x = 100x

101x = 0.1

x = 0.0009901

pOH = -log(0.0009901) = 3.00
jufsanabriasa

The pH of this solution is 12.

We can solve this question knowing that the ammonium hydroxide, NH₄OH, dissociates in water as follows:

NH₄OH(aq) ⇄ NH₄⁺(aq) + OH⁻(aq)

Based on the reaction, 1 mole of NH₄OH produces 1 mole of OH⁻

With this molarity and the 1% dissociated we can find the molarity of OH⁻. With molarity of OH⁻ we can find pOH (pOH = -log[OH⁻]) and pH (pH = 14-pOH) as follows:

Molarity OH⁻:

A solution 1.0mol dm⁻³ = 1M of NH₄OH produce 1% of OH⁻ ions because only 1% is dissociate, that is:

[tex]1M NH_4OH*(\frac{1MOH^-}{100MNH_4OH}) = 0.01M OH^-[/tex]

Now, we can find pOH as follows:

pOH:

pOH = -log [OH⁻] = 2

And pH:

pH:

pH = 14 - pOH

pH = 12

You can learn more about pH and pOH in:

https://brainly.com/question/17144456

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