Qualitative analysis of Group II(A) cations

In the systematic separation of cations we perform successive separation of group cations with the use of group reagent. Group reagents react with corresponding group cations and convert them into insoluble salts like chlorides, sulphides, and carbonates. First, we prepare a solution of the given mixture, then we add group reagent of Ist group which converts the cations of Ist group into insoluble chlorides and separate them as precipitate and then we test for IInd group cations in the filtrate of Ist group.

You have learnt in previous post that group reagent of 2^nd group is Hydrogen sulphide H₂S (gas or saturated aqueous solution).  Here you will see how cations get precipitated by common ion effect. Hydrogen sulphide is a weak acid, it dissociates partly. Hydrochloric acid, which we added initially, is a strong acid and dissociates completely.

H₂S ⇌ 2H⁺ + S^2-

HCl ⟶ H⁺ + Cl^-

Here H⁺is a common ion among H₂S and HCl. Due to complete dissociation of HCl concentration of H⁺ increases in the solution which shifts equilibrium of reaction 1 backwards. This results in precipitation of cations in the form of sulphides.

H₂S ⟶ 2H⁺ + S^2-

Take the filtrate in a boiling test tube and heat till it is nearly boiling and then pass H₂S gas under pressure in excess (30 seconds -1 min). If 2^nd group cations are present, you will get coloured precipitate of sulphides.

Black precipitate: Mercury(II) sulphide HgS, lead(II) sulphide PbS, copper(II) sulphide CuS.

Brown precipitate: Bismuth(III) sulphide Bi₂S₃, tin(II) sulphide SnS.

Yellow precipitate: Cadmium(II) sulphide CdS, arsenic(III) sulphide As₂S₃, arsenic(V) sulphide As₂S₅, tin(IV) sulphide SnS₂.

Orange precipitate: Antimony(III) sulphide Sb₂S₃, antimony(V) sulphide Sb₂S₅.

Filter the precipitate and wash with dilute hydrochloric acid HCl. The precipitate may contain IIA or IIB or both cations. To differentiate them add an excess of (5ml) yellow ammonium polysulphide (NH₄)₂S_xsolution and heat to 50-60°C for 3-4 minutes with constant stirring. sulphides of sub group IIA (Cu sub group) are insoluble in (NH₄)₂S_xwhile sulphides of sub group IIB (As sub group) dissolve in it by forming thiosalts. First we will test IIA group in the precipitate and preserve the filtrate for the test of IIB group.

Scheme for the separation of IInd Group

Wash the precipitate with small volume of dilute (1+100ml water) ammonium sulphide (NH₄)₂Ssolution then with 2% ammonium nitrate NH₄NO₃solution and reject all washings.

Among Cu sub group IIA, mercury(II) sulphide HgS is the least soluble sulphide or you can say its solubility product is lowest than others. It is insoluble in nitric acid and water.

Separation of Hg(II): Take the precipitate in a boiling test tube or beaker, add 5-10ml 2M nitric acid HNO₃, and boil gently for 2-3minutes. Black precipitate of mercury(II) sulphide HgS is obtained. Sulphides of other cations go in to the solution by forming nitrates. Filter and wash the precipitate with a little water. Keep the filtrate to test other cations of IIA.

Confirmatory test for Hg(II): Dissolve the precipitate HgS in aqua regia. Mercury(II) chloride is formed which is soluble.

            3HgS + 6HCl + 2HNO₃ ⟶ 3HgCl₂ + 3S + 2NO + 2H₂O

            Divide this solution into 3 parts.

Part 1: Add tin(II) chloride SnCl₂ solution. white silky precipitate of Mercury(I) chloride is Hg₂Cl₂ is formed.

            2Hg²⁺ + Sn²⁺ + 2Cl^- ⟶ Hg₂Cl₂ ↓ + Sn⁴⁺

            It is an example of oxidation- reduction reaction. Where Hg(II) gets reduced to Hg(I) and Sn(II) gets oxidised to Sn(IV). If more SnCl2 is added, white precipitate turns to black because of Hg(I) gets further reduced to Hg(0) metal..

           Hg₂Cl₂↓ + Sn²⁺ ⟶ Hg ↓ + Sn⁴⁺ + 2Cl^-

Part 2: Add sodium hydroxide NaOH solution in small amount, brownish-red precipitate will be obtained and on adding more NaOH, yellow precipitate of mercury(II) oxide HgO will be obtained.

            Hg²⁺ + 2OH^- ⟶ HgO ↓ + H₂O

            This reaction is the characteristic for mercury(II) ions. You can use it to differentiate Hg(II) from Hg(I). 

Part 3: Add potassium iodide KI solution slowly, red precipitate of mercury(II) iodide is formed.

             Hg²⁺ + 2I^- ⟶ HgI₂ ↓

            On adding more KI, precipitate will get dissolved by the formation of colourless tetraiodomercurate(II) ion.

            HgI₂ ↓ + 2I^- ⟶ [HgI₄]^2-

Take the filtrate, it may contain nitrates of other IIA group cations Pb(II), Bi(III), Cu(II), Cd(II). First we will separate lead Pb(II). Test a small portion of filtrate and add dilute sulphuric acid H₂SO₄ and alcohol. If white precipitate of lead sulphate PbSO₄ is obtained (less soluble in presence of alcohol) then take the remaining filtrate and add 1M sulphuric acid H₂SO₄. Concentrate in fume cupboard until white fumes appear by the decomposition of sulphuric acid. Cool it and add 10ml water, stir and allow to settle. White precipitate of lead sulphate PbSO₄ will be obtained. Filter the precipitate and keep the filtrate to test other cations.

            Pb²⁺ + SO₄^2- ⟶ PbSO₄ ↓

Confirmatory test for Pb(II): To the precipitate of lead sulphate PbSO₄ add 2ml of 6M ammonium acetate CH₃COONH₄ solution, precipitate will be dissolved by the formation of tetraacetoplumbate(II) ion.

            PbSO₄ ↓ + 4CH₃COO^- ⟶ [Pb(CH₃COO)₄]^2- + SO₄^2-

Add few drops of 2M acetic acid and then 0.1M potassium chromate K₂CrO₄solution, yellow precipitate of lead chromate PbCrO₄ will be obtained.

            Pb²⁺ + CrO₄^2- ⟶ PbCrO₄ ↓

Filtrate may contain nitrates of sulphates of Bi(III), Cu(II) and Cd(II). Add concentrated ammonia solution in excess. All of them form salts but only the salt of bismuth is insoluble in excess of ammonia. You will get white precipitate of basic salt of bismuth.

            Bi³⁺ + NO^3- + 2NH₃ + 2H₂O ⟶ Bi(OH)₂NO₃ ↓ + 2NH₄⁺

     Copper forms basic copper sulphate salt.   

            2Cu²⁺ + SO₄^2- + 2NH₃ + 2H₂O ⟶ Cu(OH)₂.CuSO₄ ↓ + 2NH₄⁺

This basic copper sulphate salt is soluble in excess of ammonia and a deep blue colouration is obtained by the formation of tetramminocuprate(II) complex ion.

            Cu(OH)₂.CuSO₄ ↓ + 8NH₃ ⟶ 2[Cu(NH₃)₄]²⁺ + SO₄^2- + 2OH^-

            Cadmium forms cadmium(II) hydroxide in ammonia solution.

            Cd²⁺ + 2NH₃ + 2H₂O ⇌Cd(OH)₂ ↓ + 2NH₄⁺

                    

Cadmium(II) hydroxide dissolves in excess of ammonia due to the formation of tetramminecadmiate(II) ion which is colourless.

                    Cd(OH)₂ ↓ + 4NH₃ ⟶ [Cd(NH₃)₄]²⁺+ 2OH^-

            You have seen that only the salt of bismuth is insoluble in excess of reagent. Filter the precipitate and test it for Bi(III) and keep the filtrate to test for remaining cations.

Confirmatory test for Bi(III): Dissolve the precipitate in a little volume of dilute hydrochloric acid HCl and pour into freshly prepared cold sodium tetrahydroxostannate(II) (2ml of 0.25M tin(II) chloride and 2ml of 2M sodium hydroxide). Black precipitate of bismuth metal will be obtained.

            Bi³⁺ + 3OH^- ⟶ Bi(OH)₃ ↓

            Sodium hydroxide present in the reagent first reacts with Bi(III) and then tetrahydroxostannate(II) ion reduces Bi(III) to Bi(0) metal.

            2Bi(OH)₃ ↓ + 3[Sn(OH)₄]^2- ⟶ 2Bi ↓ + 3[Sn(OH)₆]^2-

Take the filtrate it may contain tetramminocuprate(II) and tetramminecadmiate(II). If it is deep blue coloured then the blue colour is because of the presence of Cu(II). Otherwise test it for Cd(II). Divide the filtrate in 2 parts to test Cu(II) and Cd(II) separately.

Confirmatory test for Cu(II): take the first part and acidify it by dilute acetic acid CH3OOH and add potassium hexacyanoferrate(II)  K₂[Fe(CN)₆] solution. Reddish brown precipitate of copper hexacyanoferrate(II) will be obtained.

            Cu²⁺ + [Fe(CN)₆]^2- ⟶ Cu[Fe(CN)₆] ↓

Confirmatory test for Cd(II): Take the second part and add potassium cyanide KCN solution drop by drop until the colour disappears, then add 1 ml more. Pass hydrogen sulphide H₂S gas for 30 seconds. Yellow precipitate of CdS will be obtained.

            Cd²⁺ + 2CN^-  ⟶ Cd(CN)₂ ↓

When we add KCN white precipitate of cadmium cyanide is formed which dissolves in excess of reagent by forming tetracyanocadmiate(II) ion.

            Cd(CN)₂ ↓ + 2CN^-   ⟶ [Cd(CN)₄]^2-

This is a colourless complex ion which yields yellow coloured cadmium sulphide CdS precipitate on passing H₂S gas.

            [Cd(CN)₄]^2- + H₂S ⟶ CdS ↓ + 2H⁺ + 4CN^-

In the next post we will discuss the tests for IIB cations in the filtrate. 

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Qualitative analysis of Group I cations

In this post we will learn how to detect I^st group cations in a given mixture. In the last post, you have learnt solution preparation which is the most important step for the successful qualitative analysis.

Before we proceed for the 1^st group, we must check the presence of ammonium ion NH₄⁺ in the given mixture. Although it is 5^th group cation but it is tested first because during successive analysis a number of reagents are added some of which may contain ammonium ions and so considerable amount of ammonium ions may build up in the test solution when we reach up to 5^th group. So, it will be wise to test ammonium ion in the beginning.

Test for ammonium ion NH₄⁺: take about 1ml of original solution in a boiling test tube and add excess of sodium hydroxide NaOH and boil it gently. Ammonia gas is evolved on warming. Which can be identified by its odour, smell it after removing the test tube from the flame.

           NH₄⁺ + OH^- ⟶ NH_3(g) ↑ + H₂O

If you place red litmus paper at the mouth of test tube, ammonia turns it blue. As it is a Lewis base.

If you bring a glass rod moistened with concentrated hydrochloric acid HCl over the vapours of ammonia, white fumes of ammonium chloride are formed.

           NH₄⁺ + Cl^- ⟶ NH₄Cl_(g) ↑

With Nessler’s reagent you can perform confirmatory test for ammonium ion. Nessler’s reagent is the alkaline solution of potassium tetraiodomercurate (II) K₂HgI₄. For this test take a drop of original solution and mix it with a drop of NaOH on a watch glass. Take a drop of this mixture in a separate watch glass and add a drop of Nessler’s reagent. A brown precipitate or yellow or orange-red colouration is produced according to the amount of ammonia of ammonium ions present. This precipitate is a basic mercury(II) amido-iodine.

           NH₄⁺ + 2[HgI₄]^2- + 4OH^-  ⟶ HgO.Hg(NH₂)I_(s) ↓ + 7I^- + 3H₂O

Analysis of Group I: After detecting the presence of ammonium ion you can proceed for the group analysis. Take 15-20ml original solution in a boiling test tube or flask add an excess of dilute HCl. White precipitate indicates the presence of chlorides of Pb²⁺, Hg₂²⁺ or Ag⁺.

            Pb²⁺ + Cl^- ⟶ PbCl_2(s) ↓

           Hg₂²⁺ + 2Cl^- ⟶ Hg₂Cl_2(s) ↓

            Ag⁺ + Cl^- ⟶ AgCl_(s) ↓

But avoid large excess of HCl because lead chloride is soluble in concentrated HCl due to formation of tetrachloroplumbate(II) ion [PbCl₄]^2- is formed.

            PbCl_2(s) ↓ + 2Cl^-  ⟶ [PbCl₄]^2-

Filter the precipitate and keep filtrate for 2^nd group analysis. Wash the precipitate with 2ml of 2M HCl, and then wash it 2-3 times with 1ml cold water (because PbCl₂ is soluble in hot water) and reject washings.

Strategy to divide cations of group I: Now you have precipitate of 1^st group which may contain chlorides of Pb²⁺, Hg₂²⁺ or Ag⁺ or all. How will you identify them separately? Among them PbCl₂ is soluble in hot water but separates again in long needle like crystals on cooling so you can separate it by boiling.

Now transfer the precipitate in boiling test tube and boil with 5-10ml water and filter hot. Thus you can separate PbCl₂ from Hg₂Cl₂ and AgCl. Now you have filtrate and precipitate, in filtrate we will test for Pb²⁺ and test for Hg⁺ and Ag⁺ in precipitate.

Test for Pb(II) ion: Cool the filtrate, long needle like crystals of PbCl₂ is obtained if Pb²⁺ is present in any quantity. For confirmatory tests divide the filtrate in three parts.

Part 1: add 0.1M potassium chromate K₂CrO₄ solution. Yellow precipitate of lead chromate is obtained which is insoluble in dilute acetic acid.

            Pb²⁺ + CrO₄^2- ⟶ PbCrO₄ ↓

Part 2: Add 0.1M potassium iodide KI solution. Yellow precipitate of lead iodide PbI₂ is formed which is soluble in boiling water and deposits golden yellow plates upon cooling.

            Pb²⁺ + I^- ⟶ PbI₂ ↓

            An excess of KI dissolves the precipitate due to formation of tetraplumbate(II) ion.

            PbI₂ ↓ + 2I^- ⟶ [PbI₄]^2-

            On diluting with water the precipitate of PbI₂ reappears.

Part 3: Add dilute sulphuric acid H₂SO₄. White precipitate of lead sulphate PbSO₄ is obtained.

            Pb²⁺ + SO₄^2- ⟶ PbSO₄ ↓

Hot and concentrated sulphuric acid H₂SO₄ dissolves the precipitate by the formation of lead hydrogen sulphate Pb(HSO₄)₂.

            PbSO₄ ↓ + H₂SO₄ ⟶ Pb²⁺ + 2HSO₄^-

Lead sulphate PbSO₄ is soluble in concentrated solution of ammonium acetate CH₃COONH₄. On dissolution tetraacetoplumbate(II) is formed.

            PbSO₄ ↓ + 4CH₃COO^- ⟶ [Pb(CH₃COO)₄]^2- + SO₄^2-

How to separate Hg₂Cl₂ and AgCl: they can be separated by using ammonia solution. On reacting with ammonia, mercury(I) chloride forms insoluble complex while silver chloride forms a soluble complex.

Wash the residue (Hg₂Cl₂ and AgCl) 3-4 times with boiling water for the complete removal of PbCl₂. To ensure that add potassium chromate K₂CrO₄ solution to the washing, no precipitate indicates absence of Pb²⁺ ion. Now add 3-4ml hot dilute ammonia NH₃ solution to the residue. If Black precipitate appears, it is due to the formation of complex of Hg⁺ and collect the filtrate which may contains Ag⁺ ion.

Test for Hg(I) ion : ammonia solution converts the Hg₂Cl₂ in to a mixture of mercury(II)amidochloride and mercury metal, they both are insoluble and give black precipitate.

            Hg₂Cl₂ + 2NH₃ ⟶ Hg↓ + Hg(NH₂)Cl ↓ + NH₄⁺ + Cl^-

Mercury(II)amidochloride Hg(NH₂)Cl is a white coloured precipitate but finely divided mercury metal makes it shiny black.

Test for Ag(I) ion: Precipitate of AgCl dissolves on adding hot dilute ammonia solution due to formation of diammineargentate complex ion.

            AgCl ↓ + 2NH₃  ⇌ [Ag(NH₃)₂]+ + Cl^-

This solution should not be kept for long otherwise a precipitate of silver nitride Ag₃N (fulminating silver) is formed which explodes readily even in wet conditions. (Before disposing, add dilute nitric acid HNO₃ or hydrochloric acid which neutralises excess ammonia and prevents formation of Ag₃N.) For the confirmatory test divide this solution in two parts.

Part 1: Acidify with dilute nitric acid HNO₃, white precipitate of AgCl is obtained. When you add acid it neutralises excess ammonia and equilibrium shifts in backward direction and AgCl re-precipitates.

Part 2: Add potassium iodide KI solution, yellow precipitate of silver iodide is obtained.

            Ag⁺ + I^- ⟶ AgI ↓

Now you have learnt systematic separation and identification of group I cations. You can identify them by spot test also.

Spot test for group I cations: Wash the precipitate of group I with cold water. Add ammonia NH₃ solution. If precipitate :

Doesn’t change then Pb⁺ may present

Turns black then Hg₂²⁺ may present

Dissolves then Ag⁺ may present

Spot tests are just preliminary tests if you get positive test for any cation always do confirmatory test. In the next post we will discuss identifying tests for 2^nd group cations in the filtrate you got after the removal of precipitate of 1^st group cations. 

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Solubility and Common ion Effect

In the last post we have seen how salts are dissolved in water, how polar forces of water overcome the electrostatic attraction present between the ions and how water molecules hydrate ions to keep them separate. In this post we will study how an ionic solid dissolves in a solvent. What are the factors which govern the solubility of a solid and dissolving ability of a solvent?

In ionic solid, ions are bonded together by electrostatic force. A solvent must overcome this electrostatic attraction to split the ionic molecule. It is not so easy for a solvent to dissolve an ionic solute because it has to first break their lattice and weaken the electrostatic attraction present between the ions. Solvent also needs similar weapon to win this war. Usually solvents are covalent compounds but they may have some polar characters. If the solvent molecules are polar enough, they may weaken the electrostatic attraction between the ions of the ionic solid and then the molecules of solvent gather around the ions to solvate them, this process is called as solvation (when solvent is water we call it to hydrate the ions and the process is called Hydration). Solvation is very important because energy is released in this process and that’s how a solvent succeeds to overcome the lattice enthalpy of ionic solids.

Solvation enthalpy is a characteristic of solvent. It is lower for non polar solvents and higher for polar solvents. Lattice enthalpy must be overcome by solvation enthalpy for dissolution of a solid. That’s why non-polar substances dissolve in non-polar solvents and polar substances dissolve in polar solvents (like dissolves like). Water has higher solvation enthalpy and this is the reason why it is known as universal solvent.

On the basis of their solubility in water, solutes can be classified into three categories: insoluble, soluble and sparingly soluble. Insoluble solutes don’t dissolve in water and soluble solutes are completely soluble while the solutes which are partly dissolved fall under the category of sparingly soluble.

If you add sugar in water, it gets dissolved. If you continually add more sugar in this solution you will find that after a certain amount you won’t be able to dissolve more sugar in it. That means sugar solution gets saturated and it won’t dissolve any more sugar at this temperature and pressure. When we add more sugar in this saturated solution, an equilibrium develops between dissolved and undissolved sugar. If you want to know how you can dissolve more sugar in this saturated solution or how you can take out sugar from this solution, you will have to study this equilibrium.

Whenever a solute comes in contact with saturated solution, equilibrium is always developed between dissolved and undissolved solute. We will study this equilibrium with the example of Barium sulphate (BaSO₄):

BaSO_4(s) ⇌ Ba²⁺_(aq) + SO₄^2-_(aq)

Write the equation of equilibrium constant:

K = {[Ba²⁺][ SO₄^2-]} / [ BaSO₄]

K [BaSO₄] = [Ba²⁺] [SO₄^2-]

For a pure solid substance concentration remain constant, so we can write

K_sp = [Ba²⁺] [SO₄^2-]

K_sp is the solubility product constant of solute which is the product of molar concentrations of ions. With the help of K_sp we can calculate maximum number of moles of solute which can be dissolved in 1 L of solvent; it is called as molar solubility of a solute and it is represented by “S”. The molar concentration of ions will be equal to the molar solubility of solute. Let’s see:

On dissolution BaSO₄ dissociates into one ion of Ba²⁺ and one ion of SO₄^2-. The molar concentration of ion is equal to the molar solubility of solute.

[Ba²⁺] = S

[SO₄^2-] = S

So we can write the equation for K_sp as:

K_sp = [S] [S]

K_sp = S²

S =√K_sp

Let’s take another example of zirconium phosphate (Zr₃PO₄). On dissolution in water it dissociates in to 3 ion of Zr⁴⁺ and 4 ion of PO₄^3-.

Zr3PO4 ⇌ 3 Zr4+ + 4 PO43-

K = [Zr⁴⁺]³[PO₄^3-]⁴ / [Zr₃PO₄]

K [Zr₃PO₄] = [Zr⁴⁺]³ [PO₄^3-]⁴

K_sp = [Zr⁴⁺]³[PO₄^3-]⁴

Molar solubility of solute is equal to the molar concentration of ions. On dissolution of Zr₃PO₄three ion of Zr⁴⁺ and four ion of PO₄^3- are produced.

So the molar concentration of one Zr⁴⁺ is equal to S and for three ions it will be 3S, similarly

[Zr⁴⁺] = 3S

[PO₄^3-] = 4S

Now putting the values in the equation of K_sp

K_sp = [3S]³ [4S]⁴

K_sp = [27S]³ [256S]⁴

K_sp = 6912 S³⁺⁴

S = (K_sp/ 6912)^1/7

We can derive the general equation for salt M^a+_pX^b-_q

M^a+_pX^b-_q⇌ p M^a+ + q X^b-

K_sp = [M^a+]p [X^b-]q

On dissolution of M^a+_pX^b-_q salt p number of M^a+ions and q number of X^b- are produced. Solubility of salt is S, so the molar concentration of ions will be:

[M^a+] = [pS]

[X^b-] = [qS]

On putting the values in equation for K_sp

K_sp = [pS]^p [qS]^q

K_sp = (p^p × q^q) S^(p+q)

If the concentration of any ion is not the concentration at the time of equilibrium then K_spis called as Q_sp as we have studied in the chemical equilibrium post.

The Le Chatelier’sprinciples are also applicable to the equilibrium between saturated solution and the solute. Let’s see how we can manipulate the solubility of a solute with the help of these laws.

If we increase the concentration of one ion then the equilibrium shifts in the backward direction till Q_sp becomes equal K_sp. And thus we can precipitate salt from the solution. Similarly if we remove any ion from the solution then the equilibrium shifts in the forward direction till Q_sp = K_spand we will be able to dissolve more solute in it. In both the situations we either add or remove one of the ions from the system.

Common Ion Effect

Let’s take an example:

BaSO_4(s) ⇌ Ba²⁺_(aq)+ SO₄^2-_(aq)

If we add H₂SO₄ to the solution, it increases concentration of SO₄^2- ions and shifts equilibrium to the left side of the reaction, as a result of which Barium gets precipitated as BaSO_4(s).

H₂SO₄is a strong acid and it dissociates completely:

H₂SO_4(aq) ⟶ 2H⁺_(aq) + SO₄^2-_(aq)

SO₄^2- ion is the common ion among these two reactions thus when its concentration increases due to dissociation of sulphuric acid it shifts equilibrium to the left side of the reaction resulting in the precipitation of Barium as BaSO_4(s).

BaSO_4(s) ↓  ⥃  Ba²⁺(_aq) + SO₄^2-_(aq)

This phenomenon is known as “Common Ion Effect” and is widely used for the complete precipitation of soluble or sparingly soluble salts. We can get almost 99% pure salt by applying this phenomenon. Let’s take another example: If we want to get NaCl back from its saturated solution what will we do?

NaCl_(s) ⇌ Na⁺_(aq) + Cl^-_(aq)

If we pass HCl gas through it, concentration of Cl^- ions will be increased by dissociation of HCl and the equilibrium will be shifted in backward direction and NaCl will be precipitated.

“Common Ion Effect” is a very useful phenomenon; you will find its applications in your laboratory when you perform salt analysis

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Hydrolysis of salts

How is a salt formed? When an acid and a base react, H+ of the acid combines with OH- of the base and they neutralise each other, while remaining anion of the acid and cation of the base combine to form the salt. So we can form different categories of salts on the basis of parent acid-base:

• Salt of strong acid and strong base
• Salt of strong acid and weak base
• Salt of weak acid and strong base
• Salt of weak acid and weak base

When we put a salt in water, it dissolves in water. Polar force of water weakens the electrostatic attraction which binds the anion and cation of the salt and separates them into ions. Water molecules hydrate these ions and keep them separated. Sometimes these ions react with water molecules, anions try to polarise water molecule and if they succeed to create enough polarization it may cause the breakage of O-H bond of water. This process is called the hydrolysis.

Hydrolysis of salt

In this post we will discuss different categories of salts and see what happens when they are dissolved in water.

Let’s discuss the first category: Salt of strong acid and strong base, like NaCl. NaCl is made by the reaction of HCl and NaOH. When it is dissoved in water it dissociates to form Na⁺ and Cl^-which then react with water molecule and reproduce their parent acid and base. Since both parent acid and base are equally strong, they neutralise each other and the pH of water remains unchanged. So it may seem that such salts don’t cause hydrolysis.

NaOH + HCl ⟶ NaCl + H2O
Na⁺_(aq) + H2O ⇌ NaOH + H⁺
Cl^-_(aq)  + H⁺ ⇌ HCl

Salt of strong acid and weak base: NH4Cl is made from acid HCl and base NH₄OH. In water NH₄Cl dissociates completely in NH4⁺ and Cl^- ions. NH₄⁺ ions successfully polarize water molecule and cause hydrolysis.

NH₄OH + HCl ⟶ NH₄Cl + H₂O
NH₄⁺_(aq)  + H₂O ⟶ NH₄OH + H⁺

Ammonium ion (NH₄⁺) reacts with water and form ammonium hydroxide (NH₄OH) and H⁺. NH₄OH is a weak base which dissociates a little. Thus OH^- ions of water are consumed by NH₄⁺ and H⁺ reacts with Cl^-ions.

Cl^- _(aq) + H⁺ ⇌ HCl

HCl is a strong acid and it dissociates completely. That means H⁺ions produced by hydrolysis of NH₄Cl salt remain in solution while OH^- ions get trapped by NH₄⁺. This higher concentration of H⁺ ions makes the solution acidic.

Salt of weak acid and strong base: CH₃COONa is a salt of weak acid CH₃COOH and strong base NaOH. When it is dissolved in water it dissociates completely.  

CH₃COONa + H₂O ⟶ CH₃COO^-+ Na⁺
Na⁺ + H₂O ⇌ NaOH + H⁺
CH₃COO^- + H⁺ ⟶ CH₃COOH

Na⁺ ions cause hydrolysis and produce NaOH and H⁺ions. NaOH is a strong base and it dissociates completely thus OH^-ions of water remain in the solution. And H⁺ ions react with acetate ion (CH3COO^-) and form acetic acid, which is a weak acid. It dissociates partially so H⁺ ions of water get trapped and the pH of solution increases.

Salt of weak acid and weak base: CH₃COONH₄is a salt of weak acid CH₃COOH and weak base NH₄OH. When it is dissolved in water, it dissociates completely.  

CH₃COOH + NH₄OH ⟶ CH₃COONH₄+ H₂O
CH₃COONH₄ + H₂O ⟶ CH₃COO^-+ NH₄⁺

NH₄⁺ ions cause hydrolysis and produce weak base NH₄OH and H⁺ ions. And when H⁺ ions react with CH₃COO^- ions, they produce weak acid CH₃COOH. The resulting acid and base both are weak and dissociate partially. It means, the concentration of both H⁺ and OH^- gets affected and to calculate the pH of such solution we have to know the values of pK_aand pK_b.

pH = 7+ ½ (pK_a - pK_b)

And by using the above equation we can calculate the pH of such solutions. We have learnt how salts affect the pH of water. But not everything is completely soluble in water, there are a number of substances which are slightly soluble or insoluble in water. So what are the factors which decide the solubility of any substance? In the next post we will learn about solubility and see if it is possible to increase or decrease the solubility of any substance. 

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How to Prepare Buffer Solutions?

In the last post we have learnt that a weak acid and its conjugate base are the main constituents of a buffer solution . There are three ways to prepare a buffer solution.

1.      Weak acid and its salt
2.      Weak acid and strong base
3.      Salt of weak base and strong acid

Each of these ways will give us a pair of weak acid and its conjugate base. Let us see how.

Method I: Prepare a Buffer solution by weak acid and salt: for example we will take acetic acid (pK_{a ­­­}= 4.7) and salt sodium acetate.

CH₃COOH  ⟶  H⁺ + CH₃COO^-

Acetic acid is a weak acid so it dissociates less. It makes the acidic part of buffer.

CH₃COONa  ⟶  Na⁺ + CH₃COO^-

Sodium acetate is a salt and it completely ionizes in water and makes the basic part of buffer which is conjugate base of acetic acid.

Method II: Prepare a Buffer solution by Weak acid and strong base: for example we will take acetic acid (pK_{a ­­­}= 4.7) and NaOH.

As you have seen in the previous example, acetic acid makes the acidic part of buffer. NaOH is a strong base and it dissociates completely.

NaOH  ⟶  Na⁺ + OH^-

When it reacts with acetic acid it neutralised it and produces water and the salt sodium acetate.

CH₃COOH + Na⁺ + OH^-  ⟶  CH₃COONa + H₂O

Salt sodium acetate ionises completely in water and makes the basic part of buffer which is conjugate base of acetic acid.

Method III: Prepare a Buffer solution by Salt of weak base and strong acid: if we take sodium acetate salt and HCl let’s see what happens.

CH₃COONa  ⟶  Na⁺ + CH₃COO^-

Sodium acetate is a salt and it ionises completely in water and makes conjugate base of acetic acid.

CH₃COO^- + HCl  ⟶  CH₃COOH + Cl^-

And when HCl reacts with the conjugate base, the base accepts proton given by HCl and produces acetic acid. This makes the acidic part of buffer and so on.

You have seen that how every way reaches to the one common point. Let’s try to prepare 1M buffer of acetic acid/ sodium acetate with pH 4, pK_a of acetic acid is 4.76.

To prepare this buffer we have to find out the desired concentration of acidic form - acetic acid, and basic form - acetate ions.

By using Henderson-Hasselbalch equation:

pK_a  = pH + log [acidic form]/ [Basic form]

4.76 = 4 + log [CH₃COOH]/ [CH₃COO^-]

0.76 = log [CH₃COOH]/ [CH₃COO^-]

10^0.76 = [CH₃COOH]/ [CH₃COO^-]

5.75 = [CH₃COOH]/ [CH₃COO^-]

5.75[CH₃COO^-] = [CH₃COOH]

Concentration of acetic acid/sodium acetate buffer solution is 1M, that means:

[CH₃COOH] + [CH₃COO^-] = 1M

5.75[CH₃COO^-] + [CH₃COO^-] = 1M

6.75[CH₃COO^-] =1M

[CH₃COO^-] = 0.15M = 15mmol

So,

[CH₃COOH] = 1-0.15 = 0.85M = 85mmol

Now we have the desired concentrations of both constituents. Lets see how to prepare the buffer by each of the methods:

Methods to Prepare Buffer Solutions

I. Weak acid and its salt: For this method, we will use 85mmol of acetic acid and 15 mmol of sodium acetate to prepare buffer with pH 4.

II. Weak acid and strong base: For this method, we will need acetic acid and NaOH. When acetic acid reacts with NaOH, it gets neutralized by it and produces conjugate base

CH₃COOH + NaOH  ⟶  CH₃COONa + H₂O

CH₃COONa  ⟶  CH₃COO^- + Na⁺

That means here NaOH acts as a source of conjugate base so, the concentration of NaOH will be equal to the desired concentration of conjugate base (acetate ion). We will need 15mmol NaOH and 85mmol acetic acid to prepare this buffer.

III. Salt of weak base and strong acid: For this method, we will need sodium acetate and HCl. When they reacts:

CH₃COONa  ⟶  CH₃COO^- + Na⁺

CH₃COO^- + HCl  ⟶  CH₃COOH + Cl^-

HCl gives proton to acetate ion and produces acetic acid. Thus HCl acts as the source of acidic part of the buffer. That means we will need HCl equal to the desired concentration of acetic acid. To prepare this buffer we will need 85mmol HCl and 15mmol sodium acetate.

Now you have learnt how to prepare buffer solution by different methods. Let’s solve some problems to build better understanding.

How would you make 100 ml buffer solution with a pH 4 that is 0.3M in acetic acid and 0.2M sodium acetate using a 1M acetic acid solution and 2M CH₃COONa solution.

First we have to calculate how many moles are present in 100ml buffer solution, when concentration of acetic acid is 0.3M

M = mmol/ ml

0.3M = mmol/ 100ml

= 30mmol

Now we know that we need 30 mmol of acetic acid to prepare desired buffer. Now calculate the amount of 1M acetic acid solution to get 30mmol.

M = mmol/ ml

1M = 30mmol/ ml

= 30ml

Similarly we have to do calculations for sodium acetate. Find out the number of mmol present in 100ml buffer, when its concentration is 0.2M

M = mmol/ ml

0.2M = mmol/ 100ml

= 20mmol

Now calculate the amount of 2M sodium acetate solution to get 20mmol.

M = mmol/ ml

2M = 20mmol/ ml

= 10ml

That means to prepare desired buffer we need to mix 30ml acetic acid solution and 10 ml sodium acetate solution and make it up to 100 ml by mixing remaining 60ml water.

I hope these posts have helped you to understand the mystery of buffers but if you have any doubts, please feel free to leave a comment. 

This work is licensed under the Creative Commons Attribution-Non Commercial-No Derivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.