The Aid

An acid-base titration is a neutralization reaction that is performed in the lab in order to determine an unknown concentration of acid or base. The moles of acid will equal the moles of base at the equivalence point. Here's how to perform the calculation to find your unknown. For example, if you are titrating hydrochloric acid with sodium hydroxide:

HCl + NaOH → NaCl + H2O

You can see from the equation there is a 1:1 molar ratio between HCl and NaOH. If you know that titrating 50.00 ml of an HCl solution requires 25.00 ml of 1.00 M NaOH, you can calculate the concentration of hydrochloric acid, [HCl]. Based on the molar ratio between HCl and NaOH you know that at the equivalence point:

moles HCl = moles NaOH

MHCl x volumeHCl = MNaOH x volumeNaOH

MHCl = MNaOH x volumeNaOH / volumeHCl

MHCl = 25.00 ml x 1.00 M / 50.00 ml

MHCl = 0.50 M HCl

1. Normality (N)

Normality is equal to the gram equivalent weight of a solute per liter of solution. A gram equivalent weight or equivalent is a measure of the reactive capcity of a given molecule. Normality is the only concentration unit that is reaction dependent.

Example:

1 M sulfuric acid (H2SO4) is 2 N for acid-base reactions because each mole of sulfuric acid provides 2 moles of H+ ions. On the other hand, 1 M sulfuric acid is 1 N for sulfate precipitation, since 1 mole of sulfuric acid provides 1 mole of sulfate ions.

Dilutions

You dilute a solution whenever you add solvent to a solution. Adding solvent results in a solution of lower concentration. You can calculate the concentration of a solution following a dilution by applying this equation:

MiVi = MfVf

where M is molarity, V is volume, and the subscripts i and f refer to the initial and final values.

Example:

How many millilieters of 5.5 M NaOH are needed to prepare 300 mL of 1.2 M NaOH?

Solution:

5.5 M x V1 = 1.2 M x 0.3 L

V1 = 1.2 M x 0.3 L / 5.5 M

V1 = 0.065 L

V1 = 65 mL

So, to prepare the 1.2 M NaOH solution, you pour 65 mL of 5.5...