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 greater 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

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  

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

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.

*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.

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be

i have no clue hope you find the answer