When a weak acid and a strong base are mixed, they react according to the following net-ionic equation: HA(aq) + OH⁻(aq) → A⁻(aq) + H₂O(l). If the acid and base are equimolar, the pH of the resulting solution can be determined by considering the equilibrium reaction of A⁻ with water. If the acid is in excess, the pH can be determined from the concentrations of HA and A⁻ after the reaction. If the base is in excess, the pH can be determined from the concentration of excess OH⁻. Created by Jay.
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- I'm a bit confused. For part II, why isn't the reaction CH3COO- + H2O <=> CH3COOH + OH- used, even though it was used in part I, since there is still some CH3COO- formed from both the initial reaction and the second reaction (CH3COOH + H2O <=> H3O+ + CH3COO-)?(6 votes)
- In situation 2, wouldn't the CH3OO- react with water as well like in situation 1, making the solution more basic?(4 votes)
- Acetate, CH3OO-, does still act like a base and produces a bit of hydroxide. Both the acetic acid and acetate are engaging in an acid-base reaction in the water, but since acetic acid is in excess it will predominantly determine the pH of the solution. As Jay mentioned, we do have significant amounts of both the weak acid and its conjugate base so the solution is a buffer where both the acid and base contribute to the pH. But since we have more acid than base, the solution will be more acidic.
Hope that helps.(4 votes)
- Hello. Excuse me, for (I)-"Equal moles":
at first it is said that
CH3COOH + OH- -> CH3COO- + H2O
CH3COO- + H2O <=> CH3COOH + OH-
why once CH3COOH + OH- is an irreversible reaction but then, in the second part, it is a reversible reaction.
Also, why the reaction of CH3COOH + H2O <=> CH3COO- + H3O+ is not considered?(4 votes)
- [Instructor] Acetic acid is an example of a weak acid and sodium hydroxide is an example of a strong base. When acetic acid reacts with sodium hydroxide, an aqueous solution of sodium acetate is formed along with water. Since this reaction is an acid-base neutralization reaction, and these reactions go to completion, instead of using an equilibrium arrow, we simply draw an arrow going to the right. Next, let's write the overall ionic equation. This is also called the complete ionic equation. Let's start with acetic acid. Acetic acid is a weak acid that only partly ionizes in solution. Therefore, an aqueous solution of acetic acid, most of the acetic acid molecules stay protonated and don't turn into acetate. Therefore, we're just gonna write CH3COOH down here. We're not gonna show acetic acid ionizing into H+ and the conjugate base. However, things are different for sodium hydroxide. Sodium hydroxide is a strong base, and strong bases dissociate 100% in solution. So an aqueous solution of sodium hydroxide consists of sodium cations and hydroxide anions. Next, let's think about our products. We have an aqueous solution of sodium acetate. Sodium acetate is a soluble salt. Therefore, in aqueous solution, we would have sodium cations, Na+, and the acetate anion, CH3COO-. Because water ionizes only to an extremely small extent, we don't write it as the ions. We simply write H2O. To save some time, I've added in some aqueous subscripts and a liquid one for water, some plus signs, and the reaction arrow. So this balanced equation represents the complete ionic equation. Next, we're gonna use the complete ionic equation to write the net ionic equation for this weak acid-strong base reaction. And to do that, we first need to identify spectator ions. Remember, spectator ions don't participate in the reaction. So looking at the complete ionic equation, there are sodium cations on the left and sodium cations on the right, so we can cross out the sodium cation, that's our spectator ion. After we take out the sodium cation, what's left over is our net ionic equation. So for our net equation, we have acetic acid plus hydroxide anion, forms the acetate anion and water. Now that we have our net ionic equation, we're gonna think about three different situations. And the first situation, we have equal moles of our weak acid and our strong base. Looking at our balanced net ionic equation, the mole ratio of our weak acid to our strong base is one-to-one. Therefore, if we have equal moles of our weak acid and our strong base and the mole ratio is one-to-one, the weak and the strong base are going to completely neutralize each other and form the acetate anion. So if our goal is to figure out the pH of the resulting solution, we don't need to consider acetic acid and we don't need to consider the hydroxide ions 'cause these have been completely used up. We need to think about the acetate anion in aqueous solution. And in aqueous solution, acetate anions react with water to form acetic acid and hydroxide ions. This reaction does come to an equilibrium and since Kb is less than one, at equilibrium, there will be a large amount of reactants and only a small amount of products. However, since we are increasing the concentration of hydroxide ions in solution, the resulting solution will be basic. Therefore, at 25 degrees Celsius, the pH of the resulting solution will be greater than seven. If we wanted to calculate the actual pH, we would need some more information, but we would treat it just like a weak base equilibria problem. Also note that the hydroxide ions that we wrote down here are not the same thing as the hydroxide ions from our net ionic equation. The hydroxide ions from our net ionic equation went away completely. They were neutralized by reacting with the acetic acid. So these hydroxide ions were generated from the anion hydrolysis reaction of acetate plus water, and even though it's only a small amount or a small concentration of hydroxide ions, those are the hydroxide ions that make the pH greater than seven. For the second situation, let's think about the weak acid being in excess. Since the mole ratio of our weak acid to our strong base is one-to-one, if we have more of the weak acid than we do have the strong base, all of the strong base will be used up, and we will form the acetate anion. And when the reaction comes to completion, there'll be no more hydroxide ion in solution. So if our goal is to find the pH of the resulting solution, we don't need to consider the hydroxide ion at all since it's all been used up. We need to think about the weak acid that was in excess, acetic acid, and we also have to think about the acetate anion that was formed in the neutralization reaction. First, let's think about the weak acid that was in excess. Acetic acid will react with water to form the hydronium ion, H3O+, and the acetate anion, CH3COO-. Because the concentration of hydronium ions in solution increases, that makes the solution more acidic. Therefore, when the weak acid is in excess, the pH of the resulting solution will be less than seven at 25 degrees Celsius. If our goal is to calculate the actual pH of the resulting solution, our first thought would be, this is a weak acid equilibria problem. And it is, however, there are two sources of acetate anion. One source of acetate anion is from the reaction of acetic acid with water, and then the other source of the acetate anion came from our acid-base neutralization reaction. Therefore, one way to think about this problem is, it's actually a common-ion effect problem where the common ion is the acetate anion. The presence of a common ion decreases the ionization of acetic acid in water. However, since the concentration of hydronium ions will still increase, the pH of the solution will be less than seven. Another way to think about this problem is, when we formed the acetate anion from the neutralization reaction, we had some of the weak acid in excess. And when we have a weak acid and its conjugate base present, a buffer solution forms. And so we could also use the Henderson-Hasselbalch equation to calculate the pH of the resulting solution. For the third situation, let's think about the strong base being in excess. Since the mole ratio of our weak acid to our strong base is one-to-one, if we have the strong base in excess, the strong base is going to neutralize all of the weak acid. Therefore, when the reaction comes to completion, there will be no more weak acid present. So if our goal is to find the pH of the resulting solution, we would just have to think about the extra hydroxide ions in solution. And since there are extra hydroxide ions from our strong base, we know at 25 degrees Celsius, the pH of the solution will be greater than seven, so the resulting solution will be basic. If we wanted to calculate the actual pH of the solution, this would become a strong base pH calculation problem. And while it is true that the acetate anions that were produced from the neutralization reaction will react with water to also increase the concentration of hydroxide ions, the concentration of hydroxide ions that we would get from the anion hydrolysis of the acetate anion would be negligible to the hydroxide ions that were left over. Therefore, we could just consider the strong base to calculate the pH, and we don't need to worry about anion hydrolysis for this particular situation.