titrations

A [|titration] is an analytical term that refers to stoichiometric analysis of a reaction involving aqueous solutions. Two major reactions that are studied using titrations are acid-base reactions and precipitate reactions.

The titration of a strong acid using a strong base can be monitored using a pH probe. The graph of pH versus volume of base added to the acid can be plotted, and will resemble the graph below:

The titration of a weak acid using a strong base will result in a similar plot, but with meaningful differences, as shown in the following graph. Weak acids have a larger "buffer" zone between the starting point and the equivalence point. The equivalence point also occurs at pH greater than 7 due to the presence of a conjugate base.

Here are some basic steps to follow when dealing with a titration problem. 1. Be sure the chemical equation is balanced, and check for insoluble products. 2. Look for a starting point, where the volume, molarity, or mass of a substance is provided. 3. Look for the target (end point), a.k.a. what substance the problem is asking you to find the concentration or mass of. 4. Convert the starting material into moles. 5. Perform a mole ratio calculation. 6. Convert the target from moles into volume or mass. 7. Check for significant figures.

Most of these types of problems involve solutions, so remember to use M x V = mole.

Here is an example problem:


 * A 50.00 mL solution of silver nitrate, concentration unknown, is titrated using 75.52 mL of 0.100 M sodium chloride. What is the concentration of the silver nitrate solution?

Step 1: Write a balanced chemical equation. Also note that silver chloride is insoluble (called a precipitate).
 * AgNO3 (aq) + NaCl (aq) --> AgCl (s) + NaNO3 (aq)

Step 2: Find the starting point.
 * Both a volume and a molarity are provided for NaCl, so this is the starting point.

Step 3: Find the target.
 * The question wants you to find the concentration of the silver nitrate solution; silver nitrate is your target.

Step 4: Convert start to moles.
 * Multiply: M x V = mol, so 0.100 M x 0.07552 L = 0.007552 mol NaCl.

Step 5: Mole ratio.
 * The mole ratio between end:start is 1:1, so multiply by 1/1 = 0.007552 mol AgNO3.

Step 6: Convert to proper units.
 * To find molar concentration, divide moles by volume: 0.007552 mol / 0.05000 L = 0.15104 M AgNO3 solution.

Step 7: Significant figures.
 * The molar concentration of NaCl was 0.100 M, which has three significant figures. The answer is therefore 0.151 M AgNO3.

Try some [|online practice problems] using titrations.

Here are some more practice problems: