Quantitative and Quality Analysis

Titration is an analytical chemistry technique used to find an unknown concentration of an analyte (the titrand) by reacting it with a known volume and concentration of a standard solution (called the titrant). Titrations are typically used for acid-base reactions and redox reactions. Here's an example problem determining the concentration of an analyte in an acid-base reaction:

Titration Problem

A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl.

What was the concentration of the HCl?

Step-by-Step Solution

Step 1 - Determine [OH^-]

Every mole of NaOH will have one mole of OH^-. Therefore [OH^-] = 0.5 M.

Step 2 - Determine the number of moles of OH^-

Molarity = # of moles/volume

# of moles = Molarity x Volume

# of moles OH^- = (0.5 M)(.025 L)
# of moles OH^- = 0.0125 mol

Step 3 - Determine the number of moles of H⁺

When the base neutralizes the acid, the number of moles of H⁺ = the number of moles of OH^-. Therefore the number of moles of H⁺ = 0.0125 moles.

Step 4 - Determine the concentration of HCl

Every mole of HCl will produce one mole of H⁺, therefore the number of moles of HCl = number of moles of H⁺.

Molarity = # of moles/volume

Molarity of HCl = (0.0125 mol)/(0.050 L)
Molarity of HCl = 0.25 M

Answer

The concentration of the HCl is 0.25 M.

Another Solution Method

The above steps can be reduced to one equation

M_acidV_acid = M_baseV_base

where

M_acid = concentration of the acid
V_acid = volume of the acid
M_base = concentration of the base
V_base = volume of the base

This equation works for acid/base reactions where the mole ratio between acid and base is 1:1. If the ratio were different as in Ca(OH)₂ and HCl, the ratio would be 1 mole acid to 2 moles base. The equation would now be

M_acidV_acid = 2M_baseV_base

For the example problem, the ratio is 1:1

M_acidV_acid = M_baseV_base

M_acid(50 ml)= (0.5 M)(25 ml)
M_acid = 12.5 MmL/50 ml
M_acid = 0.25 M

Recording results

It is essential that ALL data be recorded as it is collected. This should be tabulated for easy understanding, and include inaccuracies and units. Titration is repeated until results that are within 0.1 ml are obtained. These are called concordant results. The average of the concordant results is used in calculations.

Example results table

Burette readings / cm3 ± 0.05	1st titre	2nd titre	3rd titre
Final reading	24.20	23.15	24.20
Initial reading	00.00	00.00	00.05
total volume added	24.20	23.15	23.15
average of concordant results	23.17 ± 0.1

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Question 1

In an experiment, 20.0cm3 portions of 0.065 mol dm-3NaOH were titrated against dilute HCl. The table below shows the results of the titration Burette readings {cmt] 1st 2nd 3rd Final reading [cm-] 23.50 46.60 47.40 Initial reading [cmt) 0.00 23.50 24.00 -- Volume of acid used 23.50 23.10 23.40 (a) (i) Name a suitable indicator for the titration. Give a reason for your answer.     (ii) Give the colour of the indicator in the base and at the end point.     (iii) What type of reaction is demonstrated by the experiment? [ 5 marks] (b) (i) Write a balanced equation for the reaction     (ii) Determine the average volume of acid used. [ 3 marks ] (c) Calculate the     (i) concentration of the acid in mol dm-3     (ii) concentration of the acid in g dm-3    (iii) mass of HCl in 20cm3 of solution    [H = 1.00, Cl = 35.5 ] ____________________________________________________________________________________________________ OBSERVATION The question was attempted by most candidates and the performance was good. In part (a), candidates were able to correctly name a suitable indicator for the titration with logical reason for their choice, gave the colour of the indicator in the base and at end-point and stated the type of reaction involved in the experiment as follows: (a) (i) Methyl orange/methl red/phenolphthalein Because the end point will concide with the pl+/colour change range of the indicator.Zit is a reaction between strong acid and a strong base. (ii) Colour in Base Colour at the End Point *Methyl orange Yellow Orange * Methyl red Yellow Orange * Phenolphtalein Pink Colourless (iii) Neutralization In part(b) (i) and (ii), candidates correctly wrote a balanced equation for the reaction and determined the average volume of acid used thus: (i) NaOH(aq) + HCI(aq) → NaCI(aq) + H2O(l); (ii) Average titre = 23.50 + 23.40                                              2                                  = 23.45cm3 In part (c), most candidates were able to correctly calculate the concentration of the acid in moldrrr? and gdrrr" in (i) and (ii) respectively. However, some of them lost marks because of omission/wrong units and lack of understanding of the significant figures. The expected answers from candidates were as follows: (i)        CAVA/CBVB =  1 mole ratio CA CAVA/CBVB making CA the subject CA = 1 x O. 065 x 20 correct substitution                    23.45        = 0.0554 moldm-3 Alternative Method (i) Amount of NaoH used = 0.065 x 20                                                                                                              1000                                                  = 0.0013 mol From the balanced equation of reaction: 1 mol ofNaOH required 1 mol ofHCL •• 0.0013 mol of NaOH required 0.0013 mol ofHCL i.e 23.45cm3of solution contained 0.0013 mol of HCl :. 1000cm3 of solution contained 0.0013 x 1000 of HCl                                                                                                         23.45                                                                = 0.0554 mol Hence concentration of acid = 0.0554 mol dm-3 (ii) Cone. in gdm -3 = cone in moldm-3 x molar mass        Molar mass ofHCI = 1 + 35.5        = 36.5 gmol-1        Conc. in gdm-3     = 0.0554 moldm-3 x 36.5g mol-1        = 2.02 g dm-3 In (c)(iii), candidates could not calculate the mass of HCI in 20cm3 of solution. The expected solution from candidates was a follows: 1000cm3   contains 2.02 of acid solution                 :. 20cm3   will contain               2.02 x 20                                                                          1000                                                                 = 0.0404g Alternative method  Amount of acid = 0.554x 20                                                   1000 Mass of acid = amount x molar mass                 = 0.0554 x 20 x 36.5                                 1000                                 = 0.040g