Grade 12 Chemistry Note

Volumetric Analysis

Mass Percentage:

The mass percentage of a component in a given solution is the mass of the component per 100g of the solution. For e.g.If WA is the mass of the component A, WB is the mass of the component B in a solution. Then,

 

Mass percentage of A in a solution =$\frac{{{{\rm{W}}_{\rm{A}}}}}{{{{\rm{W}}_{\rm{A}}} + {{\rm{W}}_{\rm{B}}}}}$ * 100

 

 

Example: A 10% solution of sodium chloride in water (by mass) means that 10g of sodium chloride are present in 100g of the solution.

 

Volume percentage:

This unit is used in case of a liquid dissolved in another liquid. The volume percentage is defined as the volume of the solute per 100 parts by volume of solution.

For e.g., If VA is the volume of component A present is Vsol volume of the solution.

Then,

 

Volume percentage of A in the solution =$\frac{{{{\rm{V}}_{\rm{A}}}}}{{{{\rm{V}}_{{\rm{sol}}}}}}$ * 100

 

For e.g., a 10% solution of ethanol C2H5OH, in water (by volume) means that 10cm3 of ethanol is present in 100cm3 of the solution.

Strength of a solution is defined as the amount of the solute in gm, present in one liter of the solution. It is expressed as gL-1.

Mathematically,

 

Strength =$\frac{{{\rm{Mass\: of\: solute\: in\: g}}}}{{{\rm{Volume\: of\: soln\: in\: litres}}}}$

 

Molarity:

Molarity of a solution is defined as the number of moles of solute dissolved per litre of solution.

Mathematically,

 

 

Molarity=$\frac{{{\rm{No\: of\: moles\: of\: solute}}}}{{{\rm{Volume\: of\: solution\: in\: litres\: }}}}$

 

For e.g If 'a' is the weight of the solute (in gms) present in VCC volume of the solution.

Then,

Molarity =$\frac{{\rm{a}}}{{{\rm{mol\: mass\: }}}}$ * $\frac{{1000}}{{\rm{V}}}$

Molarity is expressed by the symbol M. It can also be expressed as,

 

Molarity =$\frac{{{\rm{Strength\: in\: gms\: per\: litre}}}}{{{\rm{molecular\: mass\: of\: the\: solute}}}}$

 

Normality:

Normality of a solution is defined as the number of gram equivalents of a solute dissolved per liter of the given solution.

Mathematically it is,

 

 

Normality =$\frac{{{\rm{Number\: of\: eq}}.{\rm{\: of\: solute}}}}{{{\rm{Volume\: of\: solution\: in\: litres}}}}$

 

For e.g.If a is the weight of the solute (in gm) present in VCC volume of the solution. Then,

 

 

Normality =$\frac{{\rm{a}}}{{{\rm{eq}}.{\rm{mass\: }}}}$ *$\frac{{1000}}{{\rm{V}}}$

 

Normality is expressed by the symbol N. It can also be expressed as,

 

Normality =$\frac{{{\rm{strength\: in\: gms\: per\: litre}}}}{{{\rm{Eq\: mass\: of\: the\: solute}}}}$

 

 

 

Relationship between molarity and normality

The molarity and normality of a solution is related to each other as follows:

 

Normality = Molarity *$\frac{{{\rm{Molecular\: mass\: of\: solute}}}}{{{\rm{Eq}}.{\rm{mass\: of\: solute\: }}}}$

 

Molality:

Molality of a solution is defined as the number of moles of solute dissolved in 1000g of a solvent. Mathematically, it is expressed as,

 

Molarity = $\frac{{{\rm{Number\: of\: moles\: solute\: }}}}{{{\rm{Eq}}.{\rm{mass\: of\: solute\: }}}}$

 

Molality is expressed by the symbol m.

Molality does not change with temperature.

 

 

Formality:

In case of ionic compounds like KCl, CaCO3 etc Formality is used in place of molarity.

It is the number of gram formula masses of solute dissolved per liter of the solution. It is denoted by the symbol F. Mathematically it is given as,

 

Formality =$\frac{{{\rm{Number\: of\: Gram\: formula\: masses\: of\: solute\: }}}}{{{\rm{Volume\: of\: solution\: in\: Litres}}}}$

 

 

 

Mole Fraction:

It is the ratio of number of moles of one component (solute or solvent) to the total number of moles of all the components (solute and solvent) present in the solution. It is denoted by the symbol X. Let us suppose that a solution contains two components A and B and suppose that nA moles of A and nBmoles of B are present in the solution then,

 

 

Mole fraction of A${{\rm{X}}_{\rm{A}}}$ =${\rm{n}}\frac{{_{\rm{A}}}}{{{{\rm{n}}_{{\rm{A}} + {{\rm{n}}_{\rm{B}}}}}}}$…….. (i)

 

 

 

Mole fraction of B${{\rm{X}}_{\bf{B}}}$ =${\rm{n}}\frac{{_{\rm{A}}}}{{{{\rm{n}}_{{\rm{A}} + {{\rm{n}}_{\rm{B}}}}}}}$ …….. (ii)

 

 

 

Titration:

Selection of acid base titration:

The selection of indicators is important part of volumetric analysis. An acid – base titration is usually of four types and each sets definite criteria for selection of indicators.

- Titration involving strong acid and strong base:

In this type of reaction, there is a sharp change in pH around neutralization point. The pH of this point varies from 3.3 to 10.7. Thus, a indicator of range 3 – 10.5 can be used. Hence both phenolphthalein and methyl orange can be used.

 

-Titration involving strong acid and weak base:

In this type of titration the equivalent point lies on the pH range of 3.5 to 7.0. So, acidic indicator can be used like methyl orange.

 

- Titration involving strong base and weak acid:

The titration has equivalent point in the pH range of 7.7 to 9.7. Hence, basic indicator like phenolphthalein may be used.

 

-Titration involving weak acids and weak base.

Since, there is no sharp change in pH, accurate measurement of end point in this case is not shown by indicators.

 

Redox titration:

Redox titration (also called oxidation-reduction titration) is a type oftitration based on a redox reaction between the analyte and titrant.Redox titration may involve the use of a redox indicator and/or a potentiometer.


Example: An example of a redox titration is treating a solution of iodine with a reducing agent and using starch as indicator. Iodine forms an intensely blue complex with starch. Iodine (I2) can be reduced to iodide (I) by e.g. thiosulphate (S2O32−), and when all iodine is spent the blue color disappears. This is called aniodometric titration.

Gram equivalent weight:It is defined as the no. of parts by weight in gm of a chemical substance which combines with or is displaced by 1.008gm of Hydrogen or its equivalent, i.e. 8gm of oxygen and 35.5 gm of chlorine.

 

Equivalence point of filtration:It is defined as the theoretical point in titration equivalents of acid neutralized equivalents of base to bring complete neutralization.

 

 End points: It is defined as the point in titration, as observed by sharp change in color of indicator due to neutralization.

 

Primary standard substance: The substance whose standard solution can be prepared directly by weight is known as primary, standard substance.

 

Secondary standard solution: The substance whose standard solution can’t be prepared directly by weighing but by titration with primary standard solution are secondary standard solution.

Primary standard solution: The solution whose standard solution can be prepared by directly dissolving required amount of solute in it is called primary standard solution.

 

Indicator: An indicator is the reagent used in titration to detect the end point i.e. completion of the reaction.

 

 

PH curve:

All the following titration curves are based on both acid and alkali having a concentration of 1 mol dm-3. In each case, you start with 25 cm3of one of the solutions in the flask, and the other one in a burette. Although you normally run the acid from a burette into the alkali in a flask, you may need to know about the titration curve for adding it the other way around as well. Alternative versions of the curves have been described in most cases.

Titration curves for strong acid vs. strong base

We'll take hydrochloric acid and sodium hydroxide as typical of a strong acid and a strong base.

 

 

NaOH(Aq) + HCL(Aq)→NaCL(Aq) + H2O(L)

Running acid into the alkali

 

 

 

You can see that the pH only falls a very small amount until quite near the equivalence point. Then there is a really steep plunge. If you calculate the values, the pH falls all the way from 11.3 when you have added 24.9 cm3 to 2.7 when you have added 25.1 cm3.

Running alkali into the acid

This is very similar to the previous curve except, of course that the pH starts off low and increases as you add more sodium hydroxide solution.

 

 

 

Again, the pH doesn't change very much until you get close to the equivalence point. Then it surges upwards very steeply.

Titration curves for strong acid vs. weak base

This time we are going to use hydrochloric acid as the strong acid and ammonia solution as the weak base.

 

 

NH3(Aq) + HCL(Aq)→NH4(Aq)

 

Running acid into the alkali

 

 

 

Because you have got a weak base, the beginning of the curve is obviously going to be different. However, once you have got an excess of acid, the curve is essentially the same as before.

At the very beginning of the curve, the pH starts by falling quite quickly as the acid is added, but the curve very soon gets less steep. This is because a buffer solution is being set up - composed of the excess ammonia and the ammonium chloride being formed.

Notice that the equivalence point is now somewhat acidic (a bit less than pH 5), because pure ammonium chloride isn't neutral. However, the equivalence point still falls on the steepest bit of the curve. That will turn out to be important in choosing a suitable indicator for the titration.

Running alkali into the acid

At the beginning of this titration, you have an excess of hydrochloric acid. The shape of the curve will be the same as when you had an excess of acid at the start of a titration running sodium hydroxide solution into the acid.

It is only after the equivalence point that things become different.

A buffer solution is formed containing excess ammonia and ammonium chloride. This resists any large increase in pH - not that you would expect a very large increase anyway, because ammonia is only a weak base.

 

 

 

Titration curves for weak acid v strong base

We'll take ethanoic acid and sodium hydroxide as typical of a weak acid and a strong base.

 

 

CH3COOH(Aq) + NaOH(Aq)→CH3COONa(Aq) + H2O(L)

 

Running acid into the alkali

For the first part of the graph, you have an excess of sodium hydroxide. The curve will be exactly the same as when you add hydrochloric acid to sodium hydroxide. Once the acid is in excess, there will be a difference.

 

 

 

Past the equivalence point you have a buffer solution containing sodium ethanoate and ethanoic acid. This resists any large fall in pH.

Running alkali into the acid

 

 

 

The start of the graph shows a relatively rapid rise in pH but this slows down as a buffer solution containing ethanoic acid and sodium ethanoate is produced. Beyond the equivalence point (when the sodium hydroxide is in excess) the curve is just the same as that end of the HCl - NaOH graph.

Titration curves for weak acid v weak base

The common example of this would be ethanoic acid and ammonia.

 

CH3COOH(Aq) + NH3(Aq)→CH3COONH4(Aq)

 

It so happens that these two are both about equally weak - in that case, the equivalence point is approximately pH 7.

Running acid into the alkali

This is really just a combination of graphs you have already seen. Up to the equivalence point it is similar to the ammonia - HCl case. After the equivalence point it is like the end of the ethanoic acid - NaOH curve.

 

 

Notice that there isn't any steep bit on this graph. Instead, there is just what is known as a "point of inflexion". That lack of a steep bit means that it is difficult to do a titration of a weak acid against a weak base.

 

Derivation of normality equation:

To calculate the volume of a definite solution required to prepare solutions of other normality, the following equation is used:

 

N1V1=N2V2

 

Where,

N1=initial normality and N2=normality of the new solution,

and V1= initial volume and V2= volume of the new solution.


The normality equation is commonly used to calculate the normality of solutions after dilution. The above equation is also called as dilution formula because it helps in calculating the volume of the solvent required in diluting a concentrated solution. The same equation holds good for calculations involving molarity (M).


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