SOLUTIONS
A solution is a homogeneous mixture of two or more substances in which the proportions of the substances are identical throughout the mixture.
The major component of a solution is called the solvent and the minor components are called the solutes. In most cases water is the solvent.
The amount of solute present in a fixed quantity of solvent or solution is called the concentration of the solution. It is usually measured in grams of solute per dm^{3} of solution or in moles of solute per dm^{3} of solution. In the latter case (moldm^{3}) it is also known as the molarity of the solution.
The number of moles of solute, molarity of the solution and volume of solution can thus be related by the equation:
Number of moles = volume x molarity n = C x V

The volume of solution in this case must always be measured in dm^{3} (or litres). If the volumes are given in cm^{3} then V/1000 must be used instead.
If concentration is given in gdm^{3}, it must be converted to molarity before it can be used in the above equation. This can be done easily by dividing by the molar mass of the solute.
Concentration (gdm^{3}) = Molarity x molar mass Or C_{g} = C_{m} x m_{r}

The volume of one solution required to react with a known volume of another can be deduced from the above relationships and knowledge of the relevant chemical equation. Remember it is moles which react in the ratio shown, so all quantities must be converted to moles before the comparison can be made.
The quantitative investigation of chemical reactions by comparing reacting volumes is known as volumetric analysis. The procedure by which reacting volumes are determined is known as a titration.
In titrations, a solution whose concentration is unknown is titrated against a solution whose concentration is known. The solution of known concentration is always placed in the burette, and the solution of unknown concentration is always placed in the conical flask.
Eg 28.3 cm^{3} of a 0.10 moldm^{3} solution of NaOH was required to react with 25 cm^{3} of a solution of H_{2}SO_{4}. What was the concentration of the H_{2}SO_{4} solution?
Equation: H_{2}SO_{4} + 2NaOH —> Na_{2}SO_{4} + 2H_{2}O
Moles of NaOH = 28.3/1000 x 0.1 = 2.8 x 10^{3}
2:1 ratio so moles of H_{2}SO_{4} = 2.8 x 10^{3}/2 = 1.4 x 10^{3}
so concentration of H_{2}SO_{4} = 1.4 x 10^{3}/25 x 1000 = 0.056 moldm^{3}.
Eg Calculate the volume of 0.50 moldm^{3} nitric acid required to react completely with 5 g of lead (II) carbonate.
Equation: PbCO_{3} + 2HNO_{3} —–> Pb(NO_{3})_{2} + CO_{2} + H_{2}O
Moles of PbCO_{3} = 5/267 = 0.0187
1:2 ratio so moles of HNO_{3} = 0.0187 x 2 = 0.0375
Volume of HNO_{3} = 0.0375/0.5 x 1000 = 74.9 cm^{3}.