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🧪 ap chem


🧪 Unit 4

  •  ⏱️4 min read

4.6 Titrations - Intro and Calculations

Hope Arnett

hope arnett

Dalia Savy

dalia savy


⏱️ August 22, 2020


What Are Titrations?

Titrations are an experimental method🧪 used to analyze acid-base reactions. The image below shows a set up of the apparatus.

In the burette, there’s usually a strong acid or base🍊 with a known concentration (the titrant). In the erlenmeyer flask, there is a weak acid or base, the concentration which we are trying to find (the analyte or titrand). Both the burette and erlenmeyer flask have tick marks for measurement, so we know both solution’s initial volume. 

Along with the weak acid or base, the erlenmeyer has a drop or two of indicator. There are different types of indicators. Each indicator changes the solution’s color when the moles of acid are equal to the moles of base. 

When you carry out a titration, you gradually add acid from the burette into the flask until the indicator changes the color of the solution🔴🟣. Once this happens, the titration is at the equivalence point. A common indicator is phenolphthalein, which makes the solution light-pink at the equivalence point. Below is a sample graph of the curve:

Simple Titration Calculations

Since the equivalence point is when the number of moles are equal, we can make an equation using molarity (moles/L) and volume:

moles of acid = moles of base

Since moles = moles/Liter * Liters, we can say MaVa = MbVb M represents the concentrations of the acid and base. We know the titrant’s concentration, so we can use that for Ma . Va is the titrant’s final volume, and Vb is the analyte’s initial volume. We can plug in these values and then solve for Mb. 🧠It’s important to note that if the mole ratio of the acid and base is NOT 1:1, then you would need to multiply the appropriate sides by their mole ratio. For example, if an acid and base have a mole ratio of 1:2, then this would be the equation: MaVa = 2MbVb


The following question is from the Advanced Placement YT Channel.

  1. A solution of vinegar contains an unknown amount of acetic acid, HC2H3O2. A 25.0mL sample of vinegar is titrated with 0.650 M NaOH according to the chemical reaction below. If it requires 32.04 mL of the titrant to reach the equivalence point, what is the concentration of HC2H3O2 in the vinegar?

    1. HC2H3O2 (aq) + NaOH (aq) --> H2O (l) + NaC2H3O2 (aq)

This requires the simple use of MaVa = MbVb. Since the mole ratio is 1:1, we can just plug in and solve!🎉

(Ma)(25.0mL) =(0.650 M)(32.04mL)

Ma = 0.833 M

Acids and Bases

Acid-base reaction equations illustrate what is happening during a titration. In the Brønsted-Lowry definition, acids are proton donors and 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.

By looking at chemical equations, we can identify which species are the acid, base, conjugate acid, and conjugate base. Check out the example below:

HCl (aq) + NaOH (aq) --> NaCl (aq) + H2O (l)

Take a look at the reactants and the products. Notice that this is a double replacement reaction. The reactants’ ions split up, and the H+ from HCl joined with OH to make water. The remaining ions then combined. 

Since HCl gave up its proton, and hydroxide from NaOH accepted it, HCl is the acid, and NaOH is the base. Their corresponding products are their conjugates.

Additionally, the equation illustrates another property of acid-base reactions. When reacted completely, the acid and base neutralize each other, creating a salt and water. Water is a special case--it’s amphiprotic. This is a fancy term to describe how water can be both a proton donor and a proton acceptor. In other words, water can act as either an acid or a base.

In the first example, HCl donated a proton to water, so water acts like a base here. However, in the second equation, NaOH accepts a proton from water, so water acts like an acid. Finally, another note: ionizing acids and bases in water reveal the strengths of their conjugate pairs.


There is a quick way to figure out the conjugate acid and base! First, find the pairs!

Let's use NH3 + H2O --> NH4+ + OH- as the example.

NH3 and NH4+ are the first pair.

H2O and OH- are the second pair.

But how do we know which one is the acid and which one is the base in each pair?💭

Simple! The compound with the extra hydrogen is the acid. Therefore, NH4+ is the acid and H2O is the acid.

Review Activity

Identify the acid and base and their conjugates.

  1. H2SO4 (aq) + CH3NH2 (aq) CH3NH3+ (aq) + HSO4- (aq)

  2. NH3 (aq) + HNO2 (aq) NH4+ (aq) + NO2- (aq)


🎥 Watch: AP Chemistry - Introduction to Titrations

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🌀  Unit 3: Intermolecular Forces and Properties

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