# Titration. The aim of this investigation was to find out the accurate concentration of an acid solution, which is thought to have a concentration between 0.05 and 0.15mol dm in concentration.

Making up a standard solution

Introduction

A standard solution is one whose concentration is known exactly. Standard solutions can be prepared by weighing a mass of solid, and dissolving it with known volume of solution in a standard flask. Standard solutions can be chemically reacted with a solution of unknown concentration in order to determine the concentration of the unknown. This process of adding one solution to another solution until the reaction is just complete is known as a titration. Titration is a method of quantitative analysis, and is based on measuring volumes; it is sometimes called volumetric analysis.

Aim

The aim of this investigation was to find out the accurate concentration of an acid solution, which is thought to have a concentration between 0.05 and 0.15mol dm in concentration. To do this a titration was carried out between a week alkali and strong acid and the use of methyl orange indicator to show the end point of a titration. This is a suitable indicator for a titration between a strong acid and weak alkali.

Na2CO3(aq) + H2SO4(aq) = Na2SO4(aq) + H20(l) + CO2(g)

Apparatus

• Sulphuric acid of unknown concentration
• 2.65g of anhydrous sodium carbonate
• Methyl orange indicator
• Analytical balance
• Burette and stand
• 250cm 3 beaker
• Funnel
• Test tube
• Small beaker for waste solution
• Wash bottle and distilled water
• White tile

Making a standard solution

The sodium carbonate and sulphuric acid react in a 1:1 ratio.

The sodium carbonate solution was made to a concentration of 250cm 0.1 mol dm because this amount is between 0.05 and 0.15mol dm which is thought to be the concentration of the sulphuric acid. The next step was to calculate how much sodium carbonate was needed to make a 0.1 mol dm solution:-

First the relative atomic mass of sodium carbonate was calculated:

(Relative Atomic Masses: Na = 23, C = 12, O = 16)

Mr (Na2CO3) = (2 x 23) + 12 + (3 x 16) = 106

To make 1000cm3 of a 0.100mol dm solution of sodium carbonate you would need 10.6g of the solid. So therefore to make 250cm3 of a 0.1mol dm solution of sodium carbonate, 2.65g was needed.

Into a weighing bottle 2.65g of solid anhydrous sodium carbonate was weighed accurately and the mass was then recorded. An analytical balance was used to weigh the solid because they are very accurate. A clean beaker was rinsed with distilled water and the solid was transferred from the weighing bottle into the beaker. The weighing bottle was rinsed three times with distilled water, ...