Volumetric+Analysis+-+How+Much+Liquid

Perform an acid-base volumetric analysis

 * //In solving volumetric problems, the titration data used may be that collected by the student, or that provided by the assessor. //
 * //a titration calculation is carried out correctly. Only titre values within a range of 0.5 mL are used to calculate the average volume. The final answer must have correct units and an appropriate number of significant figures //

**QUANTITATIVE ANALYSIS ** Quantitative analysis is the process of analysing a reaction to determine the amount of reactant or product involved and hence find out the purity, concentration or formula of a substance under investigation. The analysis involves making measurements of either the mass of reactants and products (gravimetric analysis), or if the reaction is done in solution then measurements involve the concentration and volume (volumetric analysis), or concentration and absorbance (colorimetric analysis). To do such an analysis it is essential to know the formulae of the species involved and to write a correctly balanced equation which accurately reflects the molar ratio of reactants and products.

=Volumetric Analysis = In an acid-base **titration** the neutralisation reaction between an acid and base is used to determine the concentration of one of the reactants, if the concentration of the other is accurately known. Using the reaction between a solution of hydrochloric acid (unknown concentration) and sodium hydroxide (standard solution - concentration known) as an example, the practical procedure is as follows: ± 0.05 mL.
 * Rinse a conical flask with water.
 * Using a 25mL pipette a sample of the hydrochloric acid is placed in a conical flask, (an aliquot). A **pipette** is a piece of apparatus which accurately delivers a given volume -before filling it is always rinsed with a small amount of the solution.
 * A few drops of an appropriate acid-base indicator, in this case phenolphthalein, is added to the flask. The solution will remain colourless.
 * A burette is then rinsed with a sample of sodium hydroxide (the solution it is to be filled with) and then filled to just below the 0.00 mL mark. A **burette** is another piece of glassware that accurately measures out a volume of liquid.
 * The initial volume of aqueous NaOH in the burette is carefully read (to 2 decimal places - giving an accuracy of ± 0.02 mL).
 * Carefully add the NaOH to the aqueous HCl, finishing the titration as soon as the first permanent pink colour is observed. The colour change is referred to as the **end-point,** and assuming the correct indicator has been used, is the point when the acid and base have reacted in the molar ratio given by the balanced neutralisation equation.
 * By taking the difference between the initial and final volumes in the burette, the total volume of NaOH(aq) added can be carefully determined. The titration is repeated at least 3 times, or until you have obtained 3 concordant results i.e. burette volumes (titres) that, ideally, agree to within

<span style="font-family: Arial,sans-serif; font-size: 11pt;">For the titration described above it was found that an average of 18.55mL of NaOH(aq) was required to reach the endpoint of the titration. The concentration of the standard NaOH(aq) was known to be 0.0150 mol L-1. The volume of HCl (aq) used was 25.0mL.
 * <span style="font-family: Arial,sans-serif; font-size: 11pt;">Titration Calculations **
 * <span style="font-family: Arial,sans-serif; font-size: 11pt;">Example 1: **

<span style="font-family: Arial,sans-serif; font-size: 11pt;">Step 1 - Write a balanced equation for the neutralisation reaction. <span style="font-family: Arial,sans-serif; font-size: 11pt;"> NaOH(aq) + HCl (aq) ® <span style="font-family: Arial,sans-serif; font-size: 11pt;"> NaCl(aq) + H2O(l) <span style="font-family: Arial,sans-serif; font-size: 11pt;">Step 2 - Calculate the amount of standard solution (in this case NaOH(aq) ) used in the titration. <span style="font-family: Arial,sans-serif; font-size: 11pt;">Don’t forget to convert the volume measured in mL to a volume in litres, L ! <span style="font-family: Arial,sans-serif; font-size: 11pt;"> n (NaOH) = c(NaOH) x //V//(NaOH) =0.0150 mol L-1 x 0.01855 L= <span style="font-family: Arial,sans-serif; font-size: 11pt;">2.78 x 10-4 mol <span style="font-family: Arial,sans-serif; font-size: 11pt;"> ( Keep 3 sig.figs. **not 3 d.p.** through the calculations) <span style="font-family: Arial,sans-serif; font-size: 11pt;">Step 3 - Using the mole ratio in the balanced equation, calculate the number of moles of hydrochloric acid (the unknown) used in the titration. <span style="font-family: Arial,sans-serif; font-size: 11pt;"> n (NaOH) = n(HCl) i.e. ratio NaOH : HCl = 1:1 <span style="font-family: Arial,sans-serif; font-size: 11pt;">Step 4 - Using the calculated value of n(HCl) and the measured volume (expressed in litres), determine the concentration of HCl. <span style="font-family: Arial,sans-serif; font-size: 120%;"> c(HCl) = __n(HCl)__ = __2.78 x 10 -4__  = 0.0111 mol L-1 V (HCl) 0.025 mL <span style="font-family: Arial,sans-serif; font-size: 11pt;">In successive titrations of a standard solution of 0.120 mol L-1 sodium carbonate, 10.0mL of sodium carbonate was neutralised by the following volumes of hydrochloric acid of unknown concentration, 22.25mL, 20.18mL, 20.24mL and 20.15mL.
 * <span style="font-family: Arial,sans-serif; font-size: 11pt;">Example 2: **

<span style="font-family: Arial,sans-serif; font-size: 11pt;">Calculate the concentration of the hydrochloric acid; the indicator used was methyl orange.
 * <span style="font-family: Arial,sans-serif; font-size: 11pt;">Note: **<span style="font-family: Arial,sans-serif; font-size: 11pt;"> The average volume of HCl used is calculated by averaging only the **concordant** results i.e. excluding the first volume, 22.25 mL

<span style="font-family: Arial,sans-serif; font-size: 11pt;"> Na2CO3 + 2HCl ® <span style="font-family: Arial,sans-serif; font-size: 11pt;"> 2NaCl + H2O + CO2 <span style="font-family: Arial,sans-serif; font-size: 11pt;"> c = 0.120 c = ? <span style="font-family: Arial,sans-serif; font-size: 11pt;"> //V// = 10.0mL //V//(average) = 20.19 mL =0.010L= <span style="font-family: Arial,sans-serif; font-size: 11pt;">0.02019 L <span style="font-family: Arial,sans-serif; font-size: 11pt;"> n (Na2CO3) = c(Na2CO3) x //V//(Na2CO3) <span style="font-family: Arial,sans-serif; font-size: 11pt;"> = 0.120 mol L-1 x 0.010 L <span style="font-family: Arial,sans-serif; font-size: 11pt;"> = 1.20 x 10-3 mol <span style="font-family: Arial,sans-serif; font-size: 11pt;">From balanced eqn. __n(Na2CO3)__ = __n(HCl)__ <span style="font-family: Arial,sans-serif; font-size: 11pt;"> 1 2 <span style="font-family: Arial,sans-serif; font-size: 11pt;"> or n(HCl) = 2 x n (Na2CO3) =2 x 1.20 x 10-3mol= <span style="font-family: Arial,sans-serif; font-size: 11pt;">2.40 x 10-3 mol <span style="font-family: Arial,sans-serif; font-size: 11pt;">Hence concentration c(HCl) <span style="font-family: Arial,sans-serif; font-size: 15px;"> = __<span style="font-family: Arial,sans-serif; font-size: 15px;">n(HCl) __<span style="font-family: Arial,sans-serif; font-size: 15px;"> = __<span style="font-family: Arial,sans-serif; font-size: 15px;">2.4 x 10 -3 __ <span style="font-family: Arial,sans-serif; font-size: 11pt;">= **<span style="font-family: Arial,sans-serif; font-size: 11pt;">0.119 mol L-1 **

V (HCl) 0.02019 <span style="font-family: Arial,sans-serif; font-size: 11pt;"> L