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Unit 4

4.8 Intro to Acid-Base Neutralization Reactions

4 min read•august 20, 2020

Dalia Savy

AP Chemistry 🧪

269 resources

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The second main type of reaction that you learn in this unit is the acid-base neutralization reaction🍊.

Bronsted-Lowry Acid/Base

To recall, acids are proton donors while bases are proton acceptors. The species that donated the proton becomes the acid’s conjugate base. Similarly, the species that accepted the proton is now the conjugate acid.

Looking at a chemical equation, you should be able to tell what the acid-base pairs are, and picking out the conjugates.

For example🤔, H2O + H2S --> H3O+ + HS-.

First things first, what are the acid-base pairs?

First pair: H2O and H3O+

Second pair: H2S and HS-

Now, which are the acids and which are the bases. A quick way to know would be to figure out which compound in the pair has an extra hydrogen. Since H3O+ has one more hydrogen than H2O, it is the conjugate acid. This makes H2O the base.

Try the second pair on your own😊!

For more on this topic about the Bronsted-Lowry Definition:

📝Read: AP Chemistry - Titrations

There are also these weird agents🥴 called amphiprotic substances. They can both donate and accept protons! An example that you really, really know is H2O, but NH3- is also an amphiprotic substance.

💡Tip - H+ and H3O+ are used synonymously as a proton.

Neutralization

A neutralization reaction occurs when an acid and base react to often form an ionic salt and liquid water. The H+ ion from the acid combines with the OH- from the base to form H2O (l).

You often have to write out the chemical reaction. Let's say the two reactants are HNO3 (aq) and KOH (aq), what are the products💭?

To make things easier on yourself, automatically write out H2O (l)💧. Then, just combine the remaining ions, which would form the salt.

👉 HNO3 (aq) + KOH (aq) --> H2O (l) + KNO3

Soluble Salt?

Using solubility rules, is KNO3 soluble? Or is it a precipitate?

Any compound with NO3 is soluble, so KNO3 is in the aqueous state. 👉 HNO3 (aq) + KOH (aq) --> H2O (l) + KNO3 (aq)

Net Ionic Equation

📝Read: AP Chemistry - Net Ionic Equation

When writing a net ionic equation for neutralization reactions, you have to be really careful⚠️ not to dissociate weak acids and bases. Luckily, HNO3 is a strong acid and KOH is a strong base, so we can dissociate both.

📝Read: AP Chemistry - Representation of Solutions for list of strong acids and bases

Complete Ionic Equation: H+ + NO3- + K+ + OH- --> H2O (l) + K+ + NO3-

Eliminate spectator ions, which are K+ and NO3-

👉Net Ionic Equation: H+ + OH- --> H2O (l)

Concentration of Ions Pt2

With an acid-base neutralization reaction, you could also find the concentration of ions. With this chemical reaction, what are the concentrations of hydrogen ions and hydroxide ions?

💭In other words, what is [H+]? [OH-]?

Given Information: We have 0.250 M and 28.0 mL of HNO3 and 0.320 M and 53.0 mL of KOH.

Moles

Let's find the # of moles of HNO3 and KOH using:

Molarity = moles / volume in L - We have to convert the volumes we have into L by dividing by 1000.

HNO3: 0.250 M = x moles / 0.0280 L

x = 0.00700 moles of HNO3

KOH: 0.320 M = x moles / 0.0530 L

x = 0.0170 moles of KOH

Limiting Reactant

The limiting reactant is the reactant that there is less of. In this case, since there is a one to one ratio for all compounds, HNO3 is the LR.

Concentration of 0?

Since H+ is in both the limiting reactant and H2O, it has a concentration of 0. The spectator ions cannot have a concentration of 0. Remember🤔, they just help the reaction take place.

Half the question is done🥳! [H+] = 0.

[OH-]?

Finding the concentration of the excess compound is often the hardest part of the problem. First, we have to find the # of moles reacted by converting the LR into the product.

Again, since everything is one to one, the # of moles reacted is 0.00700.

Then we simply subtract from the number of excess moles we started with, which is 0.0170 moles of KOH.

0.0170 - 0.00700 = 0.010 moles unreacted.

Last but not least, we need a volume! 28.0mL + 53.00mL = 0.081 L

0.010 moles unreacted / 0.081 L = 0.12 M of OH-

Final Answer

[H+] = 0

[OH-] = 0.12

Practice, practice, practice! It's just a lottttt of stoichiometry🙃.

Net Ionic Equation Practice

Write the net ionic equation of a reaction between HNO3 and Al(OH)3.

Here are the steps you should take:

Write out the products: H2O + Al(NO3)3
Balance the equation: 3HNO3 + Al(OH)3 --> 3H2O + Al(NO3)3
Write out the states of matter using solubility rules: 3HNO3 (aq) + Al(OH)3 (s) --> 3H2O (l) + Al(NO3)3 (aq)
1. Al(OH)3 is insoluble! We cannot dissociate it in the next step
Dissociate aqueous substances: 3H+ (aq) + 3NO3 (aq) + Al(OH)3 (s) --> 3H2O (l) + Al+3 (aq) + 3NO3- (aq)
Identify spectator ions: 3H+ (aq) + 3NO3 (aq) + Al(OH)3 (s) --> 3H2O (l) + Al+3 (aq) + 3NO3- (aq)
Cross out spectator ions: 3H+ (aq) + Al(OH)3 (s) --> 3H2O (l) + Al+3 (aq)