Now realize that when we talk about strong acid-strong base titrations, they deal with stoichiometric calculations of chemical reactions involving neutralizations between strong acids and bases. We're going to ask, what is neutralization now? Neutralization is where we have a chemical reaction in which the moles of acid and base react stoichiometrically to one another. And what's important here in these types of titration or neutralization reactions is that strong acids neutralize bases. It doesn't matter if the bases are weak or strong. That's what they do in general. And strong bases neutralize acids in general. Again, it doesn't matter if they're weak or strong. In this case, we're focusing on strong acid and strong bases interacting with one another. Okay? We're not going to talk about weak acids or weak bases being involved here. That's for a later discussion. So just realize that when we're talking about a strong acid, strong base titration, all it really is is stoichiometry of acids and bases.
- 1. The Chemical World9m
- 2. Measurement and Problem Solving2h 25m
- 3. Matter and Energy2h 15m
- Classification of Matter18m
- States of Matter8m
- Physical & Chemical Changes19m
- Chemical Properties8m
- Physical Properties5m
- Temperature (Simplified)9m
- Law of Conservation of Mass5m
- Nature of Energy5m
- First Law of Thermodynamics7m
- Endothermic & Exothermic Reactions7m
- Heat Capacity16m
- Thermal Equilibrium (Simplified)8m
- Intensive vs. Extensive Properties13m
- 4. Atoms and Elements2h 33m
- The Atom (Simplified)9m
- Subatomic Particles (Simplified)12m
- Isotopes17m
- Ions (Simplified)22m
- Atomic Mass (Simplified)17m
- Periodic Table: Element Symbols6m
- Periodic Table: Classifications11m
- Periodic Table: Group Names8m
- Periodic Table: Representative Elements & Transition Metals7m
- Periodic Table: Phases (Simplified)8m
- Periodic Table: Main Group Element Charges12m
- Atomic Theory9m
- Rutherford Gold Foil Experiment9m
- 5. Molecules and Compounds1h 50m
- Law of Definite Proportions9m
- Periodic Table: Elemental Forms (Simplified)6m
- Naming Monoatomic Cations6m
- Naming Monoatomic Anions5m
- Polyatomic Ions25m
- Naming Ionic Compounds11m
- Writing Formula Units of Ionic Compounds7m
- Naming Acids18m
- Naming Binary Molecular Compounds6m
- Molecular Models4m
- Calculating Molar Mass9m
- 6. Chemical Composition1h 23m
- 7. Chemical Reactions1h 43m
- 8. Quantities in Chemical Reactions1h 16m
- 9. Electrons in Atoms and the Periodic Table2h 32m
- Wavelength and Frequency (Simplified)5m
- Electromagnetic Spectrum (Simplified)11m
- Bohr Model (Simplified)9m
- Emission Spectrum (Simplified)3m
- Electronic Structure4m
- Electronic Structure: Shells5m
- Electronic Structure: Subshells4m
- Electronic Structure: Orbitals11m
- Electronic Structure: Electron Spin3m
- Electronic Structure: Number of Electrons4m
- The Electron Configuration (Simplified)20m
- The Electron Configuration: Condensed4m
- Ions and the Octet Rule9m
- Valence Electrons of Elements (Simplified)5m
- Periodic Trend: Metallic Character4m
- Periodic Trend: Atomic Radius (Simplified)7m
- Periodic Trend: Ionization Energy (Simplified)9m
- Periodic Trend: Electron Affinity (Simplified)7m
- Electron Arrangements5m
- The Electron Configuration: Exceptions (Simplified)12m
- 10. Chemical Bonding2h 10m
- Lewis Dot Symbols (Simplified)7m
- Ionic Bonding6m
- Covalent Bonds6m
- Lewis Dot Structures: Neutral Compounds (Simplified)8m
- Bonding Preferences6m
- Multiple Bonds4m
- Lewis Dot Structures: Multiple Bonds10m
- Lewis Dot Structures: Ions (Simplified)8m
- Lewis Dot Structures: Exceptions (Simplified)12m
- Resonance Structures (Simplified)5m
- Valence Shell Electron Pair Repulsion Theory (Simplified)4m
- Electron Geometry (Simplified)7m
- Molecular Geometry (Simplified)9m
- Bond Angles (Simplified)11m
- Dipole Moment (Simplified)14m
- Molecular Polarity (Simplified)7m
- 11 Gases2h 15m
- 12. Liquids, Solids, and Intermolecular Forces1h 11m
- 13. Solutions3h 1m
- 14. Acids and Bases2h 14m
- 15. Chemical Equilibrium1h 27m
- 16. Oxidation and Reduction1h 33m
- 17. Radioactivity and Nuclear Chemistry53m
Strong Acid Strong Base Titrations (Simplified): Study with Video Lessons, Practice Problems & Examples
Strong acid-strong base titrations involve stoichiometric calculations of neutralization reactions, where moles of acid and base react in a defined ratio. In these reactions, strong acids like HCl neutralize strong bases such as barium hydroxide, producing water and a salt. To solve for unknown quantities, use a stoichiometric chart, converting volume and molarity into moles, and applying mole-to-mole comparisons based on the balanced chemical equation. This process is essential for determining molarity or mass of reactants in acid-base reactions.
Strong Acid-Base Titrations deal with stoichiometric calculations of chemical reactions involving neutralization between strong acids and bases.
Strong Acid Strong Base Titration
Strong Acid Strong Base Titrations (Simplified) Concept 1
Video transcript
Strong Acid Strong Base Titrations (Simplified) Concept 2
Video transcript
So remember, strong acid, strong base titrations are just stoichiometric questions dealing with acids and bases. Because of that, we can use a stoichiometric chart. Now remember, the chart uses the given quantity of an acid or base in this case to determine the unknown quantity of another acid or base. Here we have HCl, so hydrochloric acid, reacting with barium hydroxide to produce barium chloride and water. We're given this volume and molarity of barium hydroxide, we're given this volume of HCl, and you're looking for the missing molarity. Realize here that looking at this equation from a stoichiometric viewpoint tells us that our given amount that we start with is usually in molarity, or it can be given to us in grams, or it can be given to us in moles. Now, the molarity is usually coupled with volume. So it's usually liters of molarity. Liters of molarity leads us directly into moles of given. And remember, just like the other stoichiometric charts we've seen, once you know your moles of given, you would do the jump. Where are you going to go from moles of given to moles of unknown? And during this jump, you have to look do a mole to mole comparison, which means you have to use the coefficients in the balanced equation. Once you have your moles of unknown, you can go anywhere you want. You can use that to find the molarity of the unknown, or you can even use it to find the grams of unknown. So if you've seen my videos on stoichiometric charts before, this is kind of like a rehash, but now looking and focusing through the lens of an acid and a base undergoing a titration. Now that we've gone over the basic parts of it that we've seen before in earlier videos, move on to the next video and let's take a look at the example question.
Strong Acid Strong Base Titrations (Simplified) Example 1
Video transcript
So here it says, if it takes 25.13 ml of 0.320 molar barium hydroxide to titrate 31 ml of a solution containing hydrochloric acid, what is the molar concentration of hydrochloric acid? Alright. So to solve this, we're going to do the following steps. First, we're going to convert the given quantity into moles of given. Our given quantity is this part here. It's the volume of molarity. Think of this as a complete set. "Off" means multiply, and remember that moles equal liters times molarity. So if I change these milliliters into liters and multiply them by the molarity, that'll give me the moles of barium hydroxide.
First, I'm going to have 0.02513 liters. So I converted it into liters already. Remember that the molarity means moles over liters, so that's 0.320 moles of barium hydroxide per 1 liter. Liters cancel out, and what I've just done is found the moles of given, the moles of barium hydroxide. Now, step 2 says to do a mole-to-mole comparison to convert moles of given into moles of unknown. Alright. So we look at the balanced equation. We put moles of barium hydroxide here on the bottom, so it can cancel out, and moles of hydrochloric acid on top. According to my balanced equation, for every one mole of barium hydroxide, we have 2 moles of hydrochloric acid. So these moles cancel out. Now at this point, this will give me the moles of hydrochloric acid, which are 0.0160832 moles.
Now it says, if necessary, convert the moles of unknown into the desired units. We have to go a step further, we have to find molarity. So if the molarity is required, then divide the moles of unknown by its liters. So I just found the moles of my unknown, but that's not my molarity, that's not my molar concentration. I'm going to take those moles that I just found and then divide them by its liters. HCl has 31 ml's, which, when you convert to liters, is 0.031 liters. Here, that'll give me my molarity as 0.519 molar HCl. So, this would be my final answer for this particular question.
How many grams of HNO3 are required to completely neutralize 110.0 mL of 0.770 M LiOH?
What is the molar mass of a 0.350 g sample of a HA acid if it requires 50.0 mL of 0.440 M Sr(OH)2 to completely neutralize it? A is used as a place holder for the unknown nonmetal of the acid.
Here’s what students ask on this topic:
What is a strong acid-strong base titration?
A strong acid-strong base titration is a type of chemical reaction where a strong acid reacts with a strong base to form water and a salt. This process involves stoichiometric calculations, where the moles of acid and base react in a defined ratio. For example, hydrochloric acid (HCl) can react with barium hydroxide (Ba(OH)2) to produce barium chloride (BaCl2) and water (H2O). The goal is to determine the unknown concentration or volume of one reactant using the known values of the other reactant.
How do you calculate the molarity of an unknown solution in a titration?
To calculate the molarity of an unknown solution in a titration, you can use a stoichiometric chart. First, convert the given volume and molarity of the known solution to moles. Then, use the balanced chemical equation to find the mole-to-mole ratio between the known and unknown solutions. Finally, convert the moles of the unknown solution back to molarity using its volume. The formula is:
What is the role of a balanced chemical equation in a titration?
The balanced chemical equation is crucial in a titration because it provides the mole-to-mole ratio between the reactants. This ratio is used to convert the moles of the known reactant to the moles of the unknown reactant. For example, in the reaction between HCl and Ba(OH)2, the balanced equation is:
This tells us that 2 moles of HCl react with 1 mole of Ba(OH)2.
Why is it important to use a strong acid and a strong base in titrations?
Using a strong acid and a strong base in titrations is important because they completely dissociate in water, ensuring a clear and straightforward stoichiometric relationship. This complete dissociation allows for precise calculations and accurate determination of the unknown concentration. In contrast, weak acids and bases do not fully dissociate, complicating the stoichiometry and making the calculations less straightforward.
What is neutralization in the context of titrations?
Neutralization in the context of titrations refers to the chemical reaction where an acid and a base react stoichiometrically to form water and a salt. In a strong acid-strong base titration, this means that the moles of the acid and base react in a defined ratio, completely neutralizing each other. For example, HCl neutralizes NaOH to form NaCl and H2O. The reaction can be represented as: