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Sonja [21]
3 years ago
13

Potassium hydroxide is classified as an arrhenius base because koh contains

Chemistry
1 answer:
mixas84 [53]3 years ago
8 0

Answer: hydroxide ions

Explanation:

According to the Arrhenius concept, an acid is a substance that ionizes in the water to give hydronium ion or hydrogen ion  and a bases is a substance that ionizes in the water to give hydroxide ion .

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

As KOH can give hydroxide ions on dissociation , it is considered as arrhenius base.

KOH\rightarrow K^++OH^-

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Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base, and write the Ka expression for
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Answer:

See explanation below

Explanation:

There are several ways to know if an acid or base is strong. One method is calculating the pH. If the pH is really low, is a strong acid, and if it's really high is a strong base.

However we do not have a pH value here.

The other method is using bronsted - lowry theory. If an acid is strong, then his conjugate base is weak. Same thing with the bases.

Now, Looking at the 4 compounds, we can say that only two of them is weak and the other two are strong compounds. Let's see:

LiOH ---> Strong. If you try to dissociate :

LiOH ------> Li⁺ + OH⁻     The Li⁺ is a weak conjugate acid.

HF -----> Weak

HF --------> H⁺ + F⁻   The Fluorine is a relatively strong conjugate base.

HCl -----> Strong

This is actually one of the strongest acid.

NH₃ ------> Weak

Now writting the Ka and Kb expressions:

Ka = [H⁺] [F⁻] / [HF]

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Finally, to calculate the [OH⁻] we need to use the following expression:

Kw = [H⁻] [OH⁻]

Solving for [OH⁻] we have:

[OH⁻] = Kw / [H⁺]

Remember that the value of Kw is 1x10⁻¹⁴. So replacing:

[OH⁻] = 1x10⁻¹⁴ / 7x10⁻⁶

[OH⁻] = 1.43x10⁻⁹ M

And now, multiplying by 10¹⁰ we have:

[OH⁻] = 1.429x10⁻⁹ * 1x10¹⁰

<h2>[OH⁻] = 14.29 </h2>

Hope this helps

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