When titrating a strongmonoprotic acid and KOH at 25°C, theA) pH will be less than 7 at the equivalence point.B) pH will be grea

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When titrating a strongmonoprotic acid and KOH at 25°C, theA) pH will be less than 7 at the equivalence point.B) pH will be grea

ter than 7 at the equivalence point.C) titration will require more moles of base than acid to reach the equivalence point.D) pH will be equal to 7 at the equivalence point.E) titration will require more moles of acid than base to reach the equivalence

Chemistry

1 answer:

denis23 [38] · Answer 1 · 2022-04-16T07:27:27+03:00

Answer:

D => pH will be equal to 7 at equivalence point

Explanation:

For Strong Acid + Strong Base titrations, pH = 7 as neither ion of the salt produced will undergo hydrolysis as would weak electrolyte titrations.

BOTH CASES ARE PRESENTED FOR CONTRAST ...

Stong monoprotic acid being titrated with NaOH ...

=> HX + NaOH => NaCl + H₂O

=> NaCl => Na⁺ + Cl⁻

=> Na⁺ + H₂O => No Rxn ( formation of NaOH will not occur as a strong electrolyte prefers to remain 100% ionized)

=> X⁻ + H₂O => No Rxn (formation of HX will not occur as a strong electrolyte prefers to remain 100% ionized)

This leaves only the Auto Ionization of Water as the reaction affecting the pH of the solution at the equivalence point of a strong acid + strong base titration. That is ...

HOH ⇄ H⁺ + OH⁻ & [H⁺] = [OH⁻] = 1 x 10⁻⁷M

pH = -log[H⁺] = -log(1 x 10⁻⁷) = -(-7) = 7

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Weak Acid + Strong Base titration => pH > 7 at equivalence point

Assume => HA = weak acid

=> HA + NaOH => NaA + H₂O

=> NaA => Na⁺ + A⁻ & A⁻ is the conjugate base of a weak acid HA

=> Na⁺ + H₂O => No Rxn ( formation of NaOH will not occur as a strong electrolyte prefers to remain 100% ionized)

=> A⁻ + H₂O => HA + OH⁻ => Excess OH⁻ at equivalence pt => pH > 7

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Weak Base + Strong Acid titration => pH < 7 at equivalence point

Weak Bases => ammonia (NH₃) or ammonia derivatives (RNH₂)* in water.

(ammonia in water) => :NH₃ + H₂O => NH₄OH ⇄ NH₄⁺ + OH⁻

(ammonia derivative in water) => RN:-H₂ + H₂O => RNH₃OH ⇄ RNH₃⁺ + OH⁻

Titration of weak base with strong acid ...

=> NH₄OH + HX => NH₄X + H₂O

=> NH₄X => NH₄⁺ + X⁻

=> X⁻ + H₂O => No Rxn (formation of HX will not occur as a strong electrolyte prefers to remain 100% ionized)

=> NH₄⁺ + HOH ⇄ NH₄OH + H⁺ => Weak base is in molecular form with excess hydronium ions (H₃O⁺ = H⁺) at equivalence point => pH < 7.

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*RNH₂ is a primary amine used in the illustration, but the above process will also occur for secondary (R₂N:-H) and tertiary amines (R₃N:) in water also.