For the reaction of acetic acid, CH_3CCOH, with sodium hydroxide, NaOH. a. write

Question

For the reaction of acetic acid, CH_3CCOH, with sodium hydroxide, NaOH. a. write the complete, balanced equation. Label the acid and its conjugate base. b. write the net ionic equation. c. how many protons are there in a molecule of acetic acid? What does an acid base titration accomplish? What is an indicator and how does it function in a titration experiment? By using the proper units for volume and concentration in the titration equation, show that Moles_acid = Moles_ What is the molarity of the unknown acid if 40.5 mL of the acid can be titrated to an end point by 20.5 mL of 0.2250 M NaOH? Show your work and be careful with significant figures.

Explanation / Answer

(1)

(a) Balanced equation,

CH3COOH (aq.) + NaOH (aq.) -------------> CH3COONa (aq.) + H2O (l)

(b) Complete ionic equation:

CH3COOH (aq.) + Na+ (aq.) + OH- (aq.) ------> CH3COO- (aq.) + Na+ (aq.) + H2O (l)

Net ionic equation:

CH3COOH (aq.) + OH- (aq.) ----------> CH3COO- (aq.) + H2O (l)

Conjugate acid - base pair: CH3COOH and CH3COO-

Another conjugate acid - base pair: H2O and OH-

(c) Each acetic acid molecule can contribute one proton.

(2)

(3) For the titration of weak vs strong base, we use the indicator phenolphthalein, which changes its colour in pH range of 8.3 to 10.0.

(4) Molarity of an acid, Macid = Moles of acid / volume of acid solution in L

Macid * Vacid = Moles of acid --------- (1)

Molarity of a base, Mbase = moles of base / volume of base solution in L

Mbase * Vbase = Moles of base ----------- (2)

If we take same concetration and volume of acid and base solutions,

left sides are equal in (1) and (2), then => Moles of acid = Moles of base

(6) Neutralisation formula,

MAVA = MBVB

MA * 40.5 = 0.2250 * 20.5

MA = 0.1139 M

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