Why is the reaction of a strong acid and a strong base classified as a neutralization reaction? a. Water and a salt are produced, leaving the pH of the solution at a very high value.
b. Only water is produced, and water has a pH of 7.0.
c. Water and a salt are produced, leaving the pH of the solution at a value of 7.0.
d. Water and a salt are produced, leaving the pH of the solution at a very low value.

Answer choice C

Explanation:

Basicly for the Arrhenius and Bornsted-Lowery theories of acids & bases, acid-base reactions can be divided into 4 forms...

a. Strong Acid + Strong Base (HCl/NaOH) => pH = 7 at Eqv. Pt.

b. Weak Acid + Strong Base (HOAc/NaOH) => pH > 7 at Eqv. Pt.

c. Strong Acid + Weak Base (HCl & NH₄OH) => pH < 7 at Eqv. Pt.

d. Weak Acid + Weak Base (HOAc & NH₄OH) => pH ∝ Stronger Electrolyte

*HOAc = Acetic Acid

*NH₄OH = Ammonium Hydroxide

For each type reaction the pH at equivalence point depends upon the salt generated by the acid-base reaction. Ions of the salt, if they react with water, (hydrolysis) will shift the pH up or down depending upon which ion reacts. If there is no reaction by the salt ions then the pH will depend only upon autoionization of water which gives pH = 7. Typically Strong Acids + Strong Bases will give a pH = 7 at equivalence point because the ions of the salt will not undergo hydrolysis in water.

Example:

Strong Acid + Strong Base

HCl(aq) + NaOH(aq) => NaCl(aq) + H₂O(l)

NaCl(aq) => Na⁺(aq) + Cl⁻(aq)

Na⁺(aq) + H₂O(l) => No Rxn (theoretically NaOH, but NaOH is a strong base which prefers to remain 100% ionized in water).

Cl⁻(aq) + H₂O(l) => No Rxn (theoretically HCl, but HCl is a strong acid which prefers to remain 100% ionized in water).

Therefore, the net rxn is H⁺ + OH⁻ => H₂O & pH = 7.0

Weak Acid + Strong Base

HOAc(aq) + NaOH(aq) => NaOAc(aq) + H₂O(l)

NaOAc(aq) => Na⁺(aq) + OAc⁻(aq)

Na⁺(aq) +  H₂O(l) => No Rxn

OAc⁻(aq) +  H₂O(l) => HOAc(aq) + OH⁻(aq) => (Excess OH⁻ functions to increase pH>7 at eqv. pt.)

Strong Acid + Weak Base

HCl(aq) + NH₄OH(aq) => NH₄Cl(aq) + H₂O(l)

NH₄Cl(aq)  => NH₄⁺(aq) + Cl⁻(aq)

Cl⁻(aq) + H₂O(l) => No Rxn

NH₄⁺(aq) + H₂O(l) => NH₄OH(aq) + H⁺(aq) => (Excess H⁺ functions to decrease pH < 7 at eqv. pt. )