Volumetric Analysis

Volumetric ANALYSIS/TITRATION : Volumetric ANALYSIS/TITRATION

Introduction : Introduction Titration is a common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of a known reactant. Because volume measurements play a key role in titration, it is also known as volumetric analysis.

Definition of terms : Standard solution is a chemical term which describes a solution of known concentration. [see me for the lab, manual on standard soln.] Concentration Mass conc ( conc. In gdm-3 ) :Mass (in grams) of a substance dissolved in 1dm-3 of solution. Mathematically; Mass conc = Definition of terms Mass(g) Vol(dm3)

Definition of terms : Definition of terms Molar Conc (conc in moldm-3 ): amount of substance (in moles) present in 1dm3 of solution. Mathematically; Molar conc = Note: Molar conc a.k.a MOLARITY (M) Amount, n (mol) Volume, V(dm3)

Relationship between molar conc & mass conc : Relationship between molar conc & mass conc i.e. molar conc (M) = mass conc/molar mass Just like in solid no of mole = mass/molar mass Conc in moldm-3 = Conc in gdm-3 molar mass

Concentration of solution : Concentration of solution Concentration is just like ‘sweetness’ of a solution. Imagine: A sugar solution contains 10.0g of sugar per dm3 of solution and another contains 2.0g sugar per dm3 of solution. The more concentrated one will be sweeter. Can you identify the sweeter?

now it follows that: : now it follows that: The conc. of a solution is directly proportional to the amount(mole,n) of substance in solution at constant volume. C α n (V constant). The conc. (c) of a soln. is inversely proportional to the vol(V) of soln, if the amount(mole/mass) is constant. C α 1/v (n constant). V V V V V C C C C C

Solved problems involved concentration : Solved problems involved concentration A solution contains 2.65g of anhydrous Na2CO3 in 200cm3 of solution. Calculate the conc. of the soln in gdm-3 [Na2CO3 = 106] Hint: Do you notice that the problem is given in 2.65g per 200cm 3 ?. Good! Just express it in gdm-3 . I mean gram in 1000cm3 SIMPLE!

let’s go : let’s go Soln 200cm3 of solution contain 2.65g of Na2CO3 1000cm3 of soln will contain X X = 1000cm3 x 2.65g 200cm3 X = 13.3g Simple arithmetic! Remember 1dm3 = 1000cm3

alternatively : alternatively You can use this formula: Mass conc = mass(g)/vol(dm3) [can you remember?] Mass = 2.65g(given), Vol in dm3 = 200 1000 = 0.200dm3 .: mass conc = 2.65 0.200 =13.25gdm3

One more : One more What is the molar conc. of a solution containing 1.12g of potassium hydroxide in 250cm3 of solution? [KOH = 56] Hint: molar conc. means ?????????? Conc. in mole per dm3 Get your answer in gdm-3 and convert it to moldm-3 Then you’ve solved the problem

Now let’s do it : Now let’s do it 250cm3 contain 1.12g 1000cm3 will contain X X = 1000 x 1.12 250 X = 4.48gdm-3 Convert to molar conc. Molarity = mass conc molarmass = 4.48/56 = 0.080mol/dm3 Using formular Molar conc = amnt (mol) Vol(dm3) Mole = mass/Mm = 1.12/56 = 0.020mol Vol (dm3) = 250/1000 = 0.250dm3 Molar conc. = 0.020 0.250 = 0.080 mol/dm3

More examples : More examples What mass of sodium hydrogen trioxocarbonate (iv) NaHCO3 would be required to prepare 100cm3 of 2.0 molar solution? [NaHCO3 = 84] Remember Molar means mol/dm3 i.e. what mass is needed to prepare 2mol/dm3 You can solve it in mol & then convert it to mass. OR Convert the given mol/dm3 to gdm3 and solve the problem.

Have a look! : Have a look! 2molar soln means ????? 2mol/dm3 1000cm3 of the soln contain 2mol NaHCO3 100cm3 will contain X X = 100 X 2 = 0.2mol 1000 Convert to mass: Mass of NaHCO3 required = 0.2 X 84 = 16.8g

PRActice problems : PRActice problems

Principle of dillution (dillution factor) : Principle of dillution (dillution factor) Key Concepts The concentration of a solution is usually given in moles per dm-3 (mol dm-3 OR mol/dm3). This is also known as molarity. Concentration, or Molarity, is given the symbol C. A short way to write that the concentration of a solution of hydrochloric acid is 0.01 mol/L is to write [HCl]=0.01M The square brackets around the substance indicate concentration. The solute is the substance which dissolves. The solvent is the liquid which does the dissolving. A solution is prepared by dissolving a solute in a solvent.

Slide 17 : When a solution is diluted, more solvent is added to it, the number of moles of solute stays the same. i.e. n1 = n2 Recall, C = n ÷ V, Make n the subject and substitute, it follows that C1V1 = C2V2 where C1=original concentration of solution V1=original volume of solution C2=new concentration of solution after dilution V2=new volume of solution after dilution n1 = no of mol of solute before dilution n2 = no of mole of solute after dilution

Slide 18 : To calculate the new concentration (C2) of a solution given its new volume (V2) and its original concentration (C1) and original volume (V1). Note: V2 = V1 + vol. of water added.

Examples : Examples Calculate the new concentration (molarity) if enough water is added to 100cm3 of 0.25M sodium chloride to make up 1.5dm3. C2=(C1V1) ÷ V2 C1 = 0.25M V1 = 100cm3 = 100 ÷ 1000 = 0.100dm3 (volume must be in dm3) V2 = 1.5dm3 [NaCl(aq)]new = C2 = (0.25 x 0.100) ÷ 1.5 = 0.017M (or 0.0.017 mol/dm3)

More : More If 280cm3 of a 3moldm-3 sodium hydroxide solution is diluted to give 0.7moldm-3 soln. What is the vol. of the resulting diluted solution? What is the vol. of distilled water added to the original soln.?

Let’s do it : Let’s do it V1 = 280cm3 ,C1 = 3moldm-3 ,C2 = 0.7moldm-3 V2 = ? C1V1 = C2V2 V2 = 3 X 280 = 1200cm3 0.7 To know the vol. of distill water added V2 = V1 + vol. of distill water added. vol. of distill water added.= 1200 – 280 = 920cm3

One more! : One more! Calculate the vol. of a 12.0moldm-3 HCl that should be diluted with distilled water to obtain 1.0dm3 of a 0.05moldm-3 HCl. Soln. C1 = 12moldm-3, V1 = ? C2 = 0.05moldm-3 , V2 = 1.0dm3 I’ve done my own part, do yours!

PRActice problems : PRActice problems

Acid-Base Titrations : Acid-Base Titrations Acid-base titrations are lab procedures used to determine the concentration of a solution. We will examine it's use in determining the concentration of acid and base solutions. Titrations are important analytical tools in chemistry.

During the titration : During the titration An acid with a known concentration (a standard solution) is slowly added to a base with an unknown concentration (or vice versa). A few drops of indicator solution are added to the base. The indicator will signal, by colour change, when the base has been neutralized i.e. when [H+] = [OH-].

At the end point : At the end point At that point - called the equivalence point or end point - the titration is stopped. By knowing the volumes of acid and base used, and the concentration of the standard solution, calculations allow us to determine the concentration of the other solution.

Volumetric apparatus : Volumetric apparatus Pipette Burette Conical flask beaker

Titration Procedure : Titration Procedure Rinse 20 or 25cm3 pipette with the base solutions. Using the pipette, accurately measure 20 or 25cm3 of the base into a clean conical flask. Add 2 or 3 drops of a suitable indicator to the base in the flask. Pour the acid into the burette using a funnel. Adjust the tap to expel air bubbles and then take the initial burette reading.

Titration Procedure : Place the conical flask on a white tile under the burette. Run the solution gradually from the burette into the conical flask and swirl the flask along. Continue the addition with swirling until the end point is reached. Titration Procedure

How do you know when you are reaching the endpoint? : How do you know when you are reaching the endpoint? The indicator will begin to show a change in colour. Swirling the flask will cause the colour to disappear. ENDPOINT IS REACHED AS SOON AS THE COLOUR CHANGE IN PERMANENT. ONE DROP WILL DO IT - once the colour change has occurred, stop adding additional acid

Warning! : Warning! Do NOT continue adding until you get a deep colour change - you just want to get a permanent colour change that does not disappear upon mixing. NOTE: If a pH meter is used instead of an indicator, endpoint will be reached when there is a sudden change in pH.

Then, : Then, Record the burette reading. The difference between the final and the initial burette readings gives the volume of the acid used. The titration should be repeated two or more times and the results averaged.

Precautions during titration : Precautions during titration Rinse the burette and the pipette with the solutions to be used in them, to avoid dilution with water. The burette tap must be tight to avoid leakage. Remove the funnel from the burette before titration, to avoid an increase in the volume of the solution in the burette. CONSULT YOUR TEXTBOOKS FOR MORE PRECAUTIONS

Recording in titration : Recording in titration Titration work could be recorded thus: state the size of the pipette used in cm3 name the indicator used record your titrations in tabular form as shown below

Recording in titration : Find the average volume of acid used from any two or more titre values that do not differ by more than 0.20cm3 .This called concordancy Rough titre may be used in averaging if it is within the concordant values. Recording in titration

Indicator Selection for Titrations : Indicator Selection for Titrations

Titration Calculations : Titration Calculations Useful Information. The concentration of one of the solutions, the acid for example (CA) The volume of acid used for the titration (VA) The volume of base used for the titration (VB) What you will calculate: The concentration of the other solution, the base for example (CB)

details of the theory behind the calculations : details of the theory behind the calculations Let’s work through this example: During a titration 75.8 cm3 of a 0.100M standard solution of HCl is titrated to end point with 100.0 cm3 of a NaOH solution with an unknown concentration. What is the concentration of the NaOH solution.

The theory : The theory Begin with a balanced equation for the reaction: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) na = 1 nb = 1 (mole ratios of acid and base) Mole = concentration X volume For the acid: na = CaVa For the base: nb = CbVb na : nb (stoichiometry mole ratio) CaVa : CbVb

The Theory : The Theory na : nb CaVa : CbVb i.e. na nb CaVa CbVb Then, CaVa na CbVb nb = =

Tips on solving the problem : Tips on solving the problem Convert the given conc. (base/acid) mol/dm3 to mol/given vol(base/acid). If the conc. Is given in g/dm3, first convert to . mol/dm3 then to mol/given vol(base/acid). Use the mole ratio and mol/given vol(base/acid)., get the mol/given vol.(acid/base). Convert mol/given vol.(acid/base) to conc(acid/base). in mol/dm3 This method is called FIRST PRINCIPLE

The tips in chart : The tips in chart acid acid Molar conc. Conc. in given vol. mole ratio Conc. In given vol. molar conc. base base Mass conc. Mass conc.

Examples : Examples 20cm3 of tetraoxosulphate (vi) acid was neutralized with 25cm3 of 0.1mold-3 sodium hydroxide solution. The equation of reaction is H2SO4 + 2NaOH Na2SO4 + 2H2O Calculate (i) conc. of acid in moldm-3 (ii) mass conc. of the acid. [H=1, S= 32, O=16]

Slide 44 : Given: conc. of the base = 0.1moldm-3 Vol. of the base = 25cm3 Convert to conc. in given vol. 0.1 mol in 1000cm3 X mol in 25cm3 X = 0.1 x 25 1000 0.0025mol(per25cm3) Use mole ratio Acid : base 1 : 2 X : 0.0025 X = 0.00125mol(in given vol of the acid) i.e 20cm3 Convert to conc.(acid) in moldm-3 0.00125mol in 20cm3 X in 1000cm3 X = 0.0625mol. .: conc. of acid = 0.0625moldm-3

Example 1 continues : Example 1 continues ii mass conc. of the acid : Mass conc. = molar conc. X molar mass 0.0625 x [2+32+64] 0.0625 x 98=6.13gdm-3 Remember, always leave your answers in 3 s.f.

More : More If 18.50cm3 hydrochloric acid were neutralized by 25cm3 of potassium hydroxide solution containing 7gdm-3. what is the conc. of the acid in moldm-3? The equation of reaction: HCl + KOH HCl + H2O [K = 39, O = 16, H = 1]

Let’s solve it together : Let’s solve it together Given: Mass conc. of the base = 7gdm-3 Convert to moldm-3 : Mass conc. = 7 Molar mass [39+16+1] = 0.125 moldm-3 Mol reacted at the given vol.(25cm3) n = conc. in moldm-3 x vol.(dm3) 0.125 x 25/1000 0.003125mol Using mole ratio Acid : base 1 : 1 X : 0.003125 X = 0.003125 0.003125mol[per18.5cm3] in moldm-3 0.003125mol in 18.5cm3 X in 1000cm3 x = 0.169mol .: conc. of the acid = 0.169 moldm-3

You can use “the theory” : You can use “the theory” CaVa = na CbVb nb Example 1 again. 20cm3 of tetraoxosulphate (vi) acid was neutralized with 25cm3 of 0.1mold-3 sodium hydroxide solution. The equation of reaction is H2SO4 + 2NaOH Na2SO4 + 2H2O Calculate (i) conc. of acid in moldm-3 (ii) mass conc. of the acid. [H=1, S= 32, O=16] Cb = 0.1 moldm-3 Vb = 25cm3 Va = 20cm3 Ca = ? na = 1 nb = 2 make Ca the subject Ca = CbVb x na Va x nb Complete it, I’m tired!