Pour neutraliser 1 litre d'une solution 0,1 M( ^*) d'acide acétique CH3 CO2 H (acide faible, don

Pour neutraliser 1 litre d'une solution 0,1 M( ^*) d'acide...

Pour neutraliser 1 litre d'une solution $0,1 \mathrm{M}\left({ }^*\right)$ d'acide acétique $\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H}$ (acide faible, dont le taux de dissociation dans cette solution est seulement de $1,3 \%$ ), quelle quantité de soude faut-il utiliser : $>$ Moins de 0,1 mole?> $>$ Plus de 0,1 mole ? (*) $0,1 \mathrm{M}=0,1 \mathrm{~mol} \mathrm{~L}^{-1}$ D'après R. Meyer

Pour neutraliser 1 litre d'une solution $0,1 \mathrm{M}\left({ }^*\right)$ d'acide acétique $\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{H}$ (acide faible, dont le taux de dissociation dans cette solution est seulement de $1,3 \%$ ), quelle quantité de soude faut-il utiliser :
$>$ Moins de 0,1 mole?> $>$ Plus de 0,1 mole ?
(*) $0,1 \mathrm{M}=0,1 \mathrm{~mol} \mathrm{~L}^{-1}$
D'après R. Meyer

Key Concepts

Acid-Base Neutralization Stoichiometry

In any neutralization reaction, a strong base reacts with an acid in a defined stoichiometric ratio, typically 1:1, regardless of whether the acid is strong or weak. This means that every molecule of the acid (whether dissociated or not) is available to react with hydroxide ions. The overall reaction is governed by the total number of acid molecules, not only by those that are dissociated at equilibrium.

Weak Acids and Degree of Dissociation

Weak acids only partially ionize in solution, meaning that a small percentage of their molecules are dissociated into ions at equilibrium. However, the entire pool of acid molecules—both dissociated and undissociated—can eventually react when a strong base is added. The apparent low degree of ionization does not reduce the amount of base required for complete neutralization because the reaction with the strong base forces the remaining acid molecules to dissociate as the equilibrium shifts.

Le Chatelier’s Principle and Equilibrium Shifts

When a strong base is added to a solution of a weak acid, it reacts with the available protons, thereby disturbing the equilibrium. According to Le Chatelier’s principle, the equilibrium will shift to produce more H+ ions by further dissociating the remaining undissociated acid. This dynamic ensures that all acid molecules are eventually neutralized, requiring a stoichiometric amount of base equivalent to the total moles of acid present.

Molarity and Stoichiometric Calculations

Molarity, defined as the number of moles of solute per liter of solution, is key for calculating the amount of reactants in a neutralization reaction. In this context, a 0.1 M solution of acid contains 0.1 moles per liter. Regardless of the acid’s weak nature or low dissociation, complete neutralization involves reacting with 0.1 moles of base, as the reaction consumes the entire pool of acid molecules.

Transcript

00:02 Column 16 .83 asks how many modes of sodium acetate must be added to 2 liters of 0 .1 molar of acetic acid to give a solution that has a ph 4 .90.

00:21 So the easiest way to solve for this is to find this using the henderson -hasselback equation.

00:27 So we just want to plug in all the information we are given.

00:32 0 .90 equals the pca.

00:36 I looked up the ka from the textbook, and the ka is just the negative, the pkk is just negative volume of the ka, 1 .7 times 10 to the negative 5, plus the law of the concentration of the conjugate base over the concentration of the acid.

01:01 And we want to find the moles of the conjugate base, not the concentration.

01:08 But the good thing about this mathematical formula is that because it's a ratio, it doesn't really matter what units you use as long as the numerator and the denominator has the same units.

01:20 So in this case, it is both molar concentration units.

01:25 But if you want to find moles in the number, like in this problem, we just need to change both of them to moles over moles.

01:34 So we want to find the moles of the acid.

01:37 So they give you, we have two liters of 0 .10 molar solution of acetic acid.

01:44 So to find the moles, you just do 2 times 0 .10, which will give you two moles of the acid.

01:54 So we just want to put two in the denominator here.

01:58 And x as the modes of the sodium acetate, because that's what we're trying to solve here.

02:07 So we just want to solve for x...

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