Question

Acids are proton (H+) donors. Bases are proton (H+) acceptors. Acid-base reactions are often "reversible", meaning that they can occur in both a left-to-right and a right-to-left direction. As a result, there is one acid and one base on each side of an acid-base reaction.

1. Identify the Acids and Bases on both sides of the following reactions:

a. NH₄⁺  + OH^-  $LaTeX: \longleftrightarrow$⟷ NH₃  + H₂O

b. HCl  + H₂O  $LaTeX: \longleftrightarrow$⟷ Cl^-  + H₃O⁺

c. H₂PO₄^-  + H₂O  $LaTeX: \longleftrightarrow$⟷ HPO₄^2-  + H₃O⁺

2. Hydrochloric acid (HCl) is a "strong acid" while acetic acid found in vinegar is a "weak acid".

a. How do the terms "strong" and "weak" relate to the reversible acid-base reaction equation? (For example, reaction 1b above)

b. If one mole of HCl was placed in a liter of water, how would the amount of H₃O⁺ compare to that of a solution made from one mole of acetic acid placed in a liter of water?

c. Is it possible to make an HCl solution that has the same amount of H₃O⁺ as an acetic acid solution? Explain.

3. The level of acidity of a solution is expressed in terms of the concentration of H₃O⁺. Since this varies over such a large range, the best way to do this is using a logarithmic scale (powers of ten) known as pH.

a.  If one mole of HCl was placed in a liter of water (i.e., 1M HCl), what is the pH of the solution?

b. What is the pH of a "neutral" solution (neither acidic nor basic)?

c. What is the concentration of H₃O⁺ in a neutral solution? What is the concentration of OH^-?

Answer 1

1.

a.

$NH_{4}^{+}+OH^{-}\leftrightarrow NH_{3}+H_{2}O$

In this reaction, NH₄⁺ donates a proton and thus, it is an acid. OH^- accepts the proton and hence is a base. If the reaction is reversed, then H₂O will act as an acid donating a proton and NH₃ will act as a base by accepting the proton from H₂O.

b.

$HCl+H_{2}O \leftrightarrow Cl^{-}+H_{3}O^{+}$

In this reaction, HCl donates a proton to form Cl^-, while H₂O accepts the proton to form H₃O⁺. Hence, HCl is acid and H₂O is a base. If the reaction is reversed, then H₃O⁺ will act as an acid and Cl^- will act as a base and accepts the proton.

c.

$H_{2}PO_{4}^{-}+H_{2}O \leftrightarrow HPO_{4}^{2-}+H_{3}O^{+}$

In this reaction, H₂PO₄^- acts as an acid and H₂O acts as a base. If the reaction is reversed, then H₃O⁺ will act as a proton donor, an acid and HPO₄^2- will act as a base by accepting the proton.

2.

a.

Acid - base reactions are reversible. The strong acids ionize completely in water while weak acids do not ionize completely.

When strong acids are ionized, they are so good at donating protons that very little of the reverse reaction occurs. When weak acid ionize in water, an equilibrium is set up between the ionized and un-ionized form of the acid. The ions react very easily to reform the acid and hence a reverse reaction occurs.

b.

HCl is a strong acid and it is completely ionized in water. Hence, H₃O⁺ concentration will be equal to the concentration of the acid. While, in case of weak acid, like acetic acid, the H₃O⁺ concentration depends on the value of its dissociation constant and will be less than the concentration of acetic acid.  

Acetic acid is a weak acid, so only a small fraction of acid ionize to produce H₃O⁺ ions and acetate ions.

Thus, if one mole of HCl was placed in a liter of water, the concentration of H₃O⁺ will be one mole. While, when one mole of acetic acid is dissolved, as only small fraction ionizes, the concentration of H₃O⁺ will be less than one mole.

c.

At the same concentration, the amount of H₃O⁺ ions produced by HCl and acetic acid will not be the same.

If one mole of HCl was placed in a liter of water, it will produce one mole of H₃O⁺ ions. While, when one mole of acetic acid is dissolved, as only small fraction ionizes, the concentration of H₃O⁺ will be less than one mole.

3.

a.

pH of 1M HCl

1M HCl produces one mole H⁺.

pH = - log[H⁺]

pH = - log(1.0) = 0

b.

The pH of the neutral solution is 7.

c.

The pH of the neutral solution is 7.

Hence, [H₃O⁺] = 10^-pH = 10^-7 = 1*10^-7M

pH + pOH = 14

Hence, [OH^-] = 1*10^-7M