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.

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole