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For this discussion, please address the following: Describe how you personally experience acid/base chemistry in your...

For this discussion, please address the following:

  1. Describe how you personally experience acid/base chemistry in your own daily life. The context could be nutritional, medicinal, agricultural, industrial, commercial, or any other relevant application of your choice. Are neutralization reactions reversible? Fully explain any formulas/calculations that you find
  2. How do industrial and vehicular emissions contribute to the ongoing problem of acid rain? Describe how acidification of water may be damaging to buildings, forests, or human health.
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Answer #1

In everyday life, the role of acids and bases is very significant. Without acid and bases, many of household work and many biological activities cannot be imagined.

When we start our day, we use a base to clean our teeth, Toothpaste contains a base which is actually a flouride base, NaF(Sodium Flouride), as the bacteria which are present in mouth are a bit acidic, hence to neutralize their effects toothpaste are basic.

Many of us take lemon tea, orange juice etc in breakfast, which is acidic in nature, whereas the potatoes are having basic components.

Antacids are the medicine used when someone is suffering from acidity, is actually a base which neutralizes the acid in the stomach.

In agriculture also, the acids and the bases play important role in maintaining the pH of the soil.

The soil in the region of heavy rainfall becomes acidic, and the pH is maintained by adding pulverized dolomitic or calcitic lime.

The soap and the detergents which are cleaning agents and used for the cleaning purposes, also are bases and remove the dirt.

Some of the acid-base reaction is reversible, the reaction between a weak acid and weak base are reversible, as the weak acid and weak base are not completely dissociated in water.

CH3COOH ⇌  CH3COO- + H+

NH4OH  ⇌ NH4+ + -OH

As the weak acid/base forms salts they will be dissociated and will be in equilibrium as shown below:

CH3COOH +  NH4OH ------> CH3COONH4 + H2O

CH3COONH4 + H2O  ⇌ CH3COO- + H+ +  NH4+ + -OH

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