# Titration Research Paper

3106 WordsDec 8, 201213 Pages
Quantitative Chemistry –Titration Determination of the Molarity of an Unknown Solution through Acid-Base Titration Technique 1. Introduction 1.1 Aim The aim of this investigation was to determine the precise molarity of two (NaOH(aq)) sodium hydroxide solutions produced at the beginning of the experiment through the acid-base titration technique. 1.2 Theoretical Background Titration is a method commonly used in laboratory investigations to carry out chemical analysis. The most frequent chemical analysis performed through titration is when determining the exact concentration of a solution of unknown molarity. This technique is usually used in redox and acid-base reactions. Redox reaction is when reduction –…show more content…
Therefore, after the titration was performed as explained on the previous paragraphs, the data needed to calculate the molarity of NaOH(aq) was obtained. 1.3 Preliminary calculations 1.3.1 The first important value to be obtained from the investigation was the volume of NaOH(aq) used. This was done by the following equation: eq.3 – for 1st solution produced Average volume volume of 2nd trial - volume of 1st trial2= V1 eq.4 – for 2nd solution produced Average volume volume of 2nd trial - volume of 1st trial2= V2 1.3.2 The next step when determining the molarity of NaOH(aq) was to calculate the moles of HCl(aq) by using the volume HCl(aq) provided on the lab scripts and the molarity obtained from the bottle of HCl(aq) used during the investigation. The eq.5 and eq.6 below was used to calculate: eq.5 – moles1 = V1 (dm3) × molarity (M) eq.6 – moles2 = V2 (dm3) × molarity (M) 1.3.3 The third important equation, for both solutions, worth noting are the number of moles of NaOH(aq) present in the reaction. This was obtained by using ratio of the moles of NaOH(aq) : HCl(aq) used during the investigation. This can be recalled by eq.2 eq.2 – HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) 1 : 1 1.3.4 Hence, moles of both solutions of NaOH(aq) will be the same as the number of moles of HCl(aq) since the mole ratio is 1:1. That is for every one mole of HCl(aq)