{"id":6083,"date":"2020-10-13T14:09:01","date_gmt":"2020-10-13T08:39:01","guid":{"rendered":"http:\/\/astan.lk\/al_virtualclassroom\/?p=6083"},"modified":"2020-10-13T14:09:50","modified_gmt":"2020-10-13T08:39:50","slug":"acidic-basic-salt-compounds","status":"publish","type":"post","link":"https:\/\/astan.lk\/al_virtualclassroom\/acidic-basic-salt-compounds\/","title":{"rendered":"Acidic, basic, salt compounds"},"content":{"rendered":"

Ionisation of water and Ionic Product of Water ( K<\/span>w<\/span><\/sub>)<\/h3>\n

Pure water being a weak electrolyte under goes self ionization to a small extent as follows:<\/span><\/p>\n

\"\"<\/span><\/p>\n

The equilibrium constant for this reaction is: \"\"<\/span><\/p>\n

The concentration of unionized water is taken as constant because the degree on ionization of water is very small. So we can write this equation as:<\/span><\/p>\n

\"formula<\/span><\/h5>\n

where Kw<\/sub> is a constant and is known as the ionic product of water whose value is 1.008 x 10<\/span>–<\/span><\/sup>1<\/span><\/sup>4<\/span><\/sup> mol<\/span>2<\/span><\/sup>\u00a0dm<\/span>–<\/span><\/sup>6\u00a0<\/span><\/sup>at 298 K. In pure water the concentration of H<\/span>3<\/span><\/sub>O<\/span>+<\/span><\/sup> and OH<\/span>–<\/span><\/sup> are equal and so we can write,<\/span><\/p>\n

[H<\/span>3<\/span><\/sub>O<\/span>+<\/span><\/sup>] = [OH<\/span>–<\/span><\/sup>]<\/span><\/p>\n

If, K<\/span>w<\/span><\/sub> = [H<\/span>3<\/span><\/sub>O<\/span>+<\/span><\/sup>] [OH<\/span>–<\/span><\/sup>] = 1.008 x 10<\/span>-14<\/span><\/sup> mol<\/span>2<\/span><\/sup> \u00a0dm–<\/sup>6<\/sup><\/span>\u00a0then,<\/span><\/p>\n

[H<\/span>3<\/span><\/sub>O<\/span>+<\/span><\/sup>] [OH<\/span>–<\/span><\/sup>] = 1.008 x 10<\/span>-14<\/span><\/sup><\/p>\n

[H<\/span>3<\/span><\/sub>O<\/span>+<\/span><\/sup>]<\/span>2<\/span><\/sup> = 1.008 x 10<\/span>-14<\/span><\/sup><\/p>\n

\"value<\/span><\/h5>\n

Thus in pure water [H<\/span>3<\/span><\/sub>O<\/span>+<\/span><\/sup>] = [OH<\/span>–<\/span><\/sup>] = 1.0 x 10<\/span>–<\/span><\/sup>7<\/span><\/sup> mol \u00a0dm–<\/sup>3<\/sup><\/span>\u00a0at 298 K<\/span><\/p>\n

<\/div>\n

Effect of temperature on K<\/span><\/h4>\n

The value of Kw varies with the change in temperature. The values of [H<\/span>3<\/span><\/sub>O<\/span>+<\/span><\/sup>] and [OH<\/span>–<\/span><\/sup>] are always equal to each other at all temperatures but the values of K<\/span>w<\/span><\/sub> are different at different temperatures. The value of K<\/span>w<\/span><\/sub>increases with the rise in temperature. This is because increase in temperature will shift the equilibrium in the forward direction producing large concentrations of [H<\/span>3<\/span><\/sub>O<\/span>+<\/span><\/sup>] and [OH<\/span>–<\/span><\/sup>] ions (Le Chatelier’s principle).<\/span><\/p>\n

\"\"<\/span><\/p>\n

Hence, K<\/span>w<\/span><\/sub> increases with rise in temperature.<\/span><\/p>\n

 <\/p>\n

It can be concluded that the hydronium and hydroxyl ions are always present in solution whether they are acidic or basic. However their concentrations differ.<\/p>\n

<\/p>\n

 <\/p>\n

Dissociation\/Ionisation constant of a weak acid (Ka<\/sub>)<\/h3>\n

An acid HA takes part in the following proton transfer equilibrium in water<\/p>\n

HA(aq) + H2<\/sub>O(l) \u00a0\u21cc \u00a0<\/span><\/span><\/span><\/span><\/span><\/b>H3O+<\/sup>(aq) + A–<\/sup>(aq)<\/p>\n

 <\/p>\n

In this expression, A–<\/sup> is the conjugate base of the acid.\"\"<\/a><\/p>\n

In dilute solution, the concentration of water is constant and the equilibrium can be expressed in terms of the equilibrium constant, Ka<\/sub>.<\/p>\n

\"\"<\/a><\/p>\n

Ka<\/sub> is the ionisation constant or dissociation constant of the acid.<\/p>\n

For acetic acid<\/p>\n

<\/p>\n

\"\"<\/a><\/p>\n

Ka is a measure of the acidic strength of the acid.<\/p>\n

 <\/p>\n

Dissociation\/Ionisation constant of a weak base (Kb<\/sub>)<\/h3>\n

\u2022 For a base B in water, the characteristic proton transfer equilibrium is<\/p>\n

B(aq) + H2<\/sub>O(l) \u00a0 \u21cc \u00a0<\/span><\/span><\/span><\/span><\/span><\/b>\u00a0<\/span><\/b>BH+<\/sup>(aq)<\/span>\u00a0+ OH–<\/sup>(aq)<\/span><\/span><\/span><\/p>\n

\"\"<\/a><\/p>\n

Here, BH+<\/sup> is the conjugate acid. In dilute solutions concentration of H2<\/sub>O is constant.<\/p>\n

\"\"<\/a><\/p>\n

Kb is the ionisation or dissociation constant of the base. It is a measure of the strength of the base.<\/p>\n

For Ammonia,<\/p>\n

NH3<\/sub>(aq) + H2<\/sub>O (l)\u00a0 \u21cc \u00a0NH4<\/sub>+<\/sup>(aq)<\/span><\/span><\/span><\/span><\/span><\/b><\/strong>\u00a0+ OH–<\/sup>(aq)<\/p>\n

Kb<\/sub> = ( [NH4<\/sub>+<\/sup>] [OH\u00af] ) \/ [NH3<\/sub>]<\/p>\n

Kb is a measure of the strength of the base.<\/p>\n

 <\/p>\n

Relationship between Ka and Kb<\/h4>\n

Let BH+<\/sup> be the conjugate acid of a base, then the expression for the acidic constant K<\/i>a<\/sub> for the conjugate acid:<\/p>\n

BH+<\/sup><\/span>\u00a0\u00a0 \u21cc \u00a0<\/span><\/span><\/span><\/span><\/span><\/b><\/strong>B<\/span> + H+<\/sup><\/dir>can be written as<\/p>\n
          [B<\/span>] [H+<\/sup>]\r\n  K<\/i>a<\/sub>  =  ---------\r\n           [BH+<\/sup><\/span>]\r\n\r\n          [B<\/span>] [H+<\/sup>]  [OH-<\/sup>]<\/span>\r\n      =  --------- ------\r\n           [BH+<\/sup><\/span>]    [OH-<\/sup>]<\/span>\r\n\r\n           [B<\/span>]\r\n      =  -----------  [H+<\/sup>] [OH-<\/sup>]<\/span>\r\n         [BH+<\/sup><\/span>] [OH-<\/sup>]\r\n\r\n          1\r\n      =  --- K<\/i>w<\/sub><\/span>\r\n          K<\/i>b<\/sub>\r\n\r\n<\/pre>\n
Thus,\u00a0K<\/i>a<\/sub><\/span> K<\/i>b<\/sub><\/span> = K<\/i>w<\/sub><\/span><\/p>\n
Furthermore,<\/p>\n
– log ( K<\/i>a<\/sub><\/span>) – log (K<\/i>b<\/sub><\/span>) = -log (K<\/i>w<\/sub><\/span>)<\/p>\n
and at 298 K, we have<\/p>\n
p K<\/i>a<\/sub><\/span> + pK<\/i>b<\/sub><\/span> = 14<\/span><\/p>\n
 <\/p>\n
\u2022 Weak acids and weak bases partially dissociate in water leading to equilibria<\/p>\n
\"\"<\/a><\/p>\n
\u2022 Strong acids like HCl, HNO3<\/sub> and strong bases like NaOH completely ionise in water.<\/p>\n
\"\"<\/a>Therefore, the H+<\/sup> and OH–<\/sup> concentration can be directly computed considering the
\ndissociation.<\/p>\n
<\/h3>\n
Ostwald’s law of dilution<\/h3>\n
For a weak acid\u00a0<\/span><\/h4>\n
An acid of the type HA can undergo ionization when dissolved in water as,<\/span><\/p>\n
\"\"<\/span><\/p>\n
If ‘n’ moles of the acid are dissolved in ‘V’ units of volume (litres L) and ‘<\/span>a<\/span><\/span>‘ is the degree of ionization, then the equilibrium amounts of various species and the concentration in moles per litre in the solution are,<\/span><\/p>\n
Note that 1 L = 1 dm3<\/sup><\/p>\n
\"\"<\/span><\/p>\n
\"\"<\/span><\/p>\n
where ‘C’ is the molar concentration of the acid. The ionization constant for the above reaction is given by,<\/span><\/p>\n
\"\"<\/span><\/p>\n
\"\"<\/span><\/p>\n
\"\"<\/span><\/p>\n
where ‘V’ is the volume of the solution in litres containing one mole of the acid HA. As the degree of ionization increases with-dilution then, the hydronium ion or hydrogen ion concentration is given by,<\/span><\/p>\n
\"deriavtion<\/span><\/h5>\n
 <\/p>\n
For a weak base<\/h4>\n
The ionization of a weak base is characterized by the equilibrium,<\/p>\n
<\/p>\n
If ‘n’ moles of the base are dissolved in ‘V’ units of volume (litres L) and <\/span>a<\/span><\/span> is the degree of ionization, then the equilibrium amounts of various species and the concentration in moles per litre in the solution are,<\/span><\/p>\n
<\/span><\/p>\n
where ‘C’ is the molar concentration of the base. The ionization constant for the above reaction is given by,<\/span><\/p>\n
<\/span><\/p>\n
If ‘\u03b1<\/span>‘ is small then 1 – \u03b1<\/span>\u00a0\u2248<\/span><\/span> 1<\/span><\/p>\n
Hence, Kb = C\u03b1<\/span>2<\/span><\/sup><\/p>\n
<\/span><\/p>\n
where ‘V’ is the volume of the solution in litres containing one mole of the base MOH. As the degree of ionization increases with dilution then, the hydroxide concentration is given by,<\/span><\/p>\n
\"derivation<\/span><\/h5>\n
 <\/p>\n
pH Value<\/h4>\n
\u2022 The concentration of hydronium ions [H3<\/sub>O+<\/sup>] in aqueous solutions is sometimes very small. Hydronium ion concentration of a medium is very important in industrial processes, pollution, acid rain, blood and body chemistry. In many cases its concentration ranges from 10-14<\/sup> to 10 mol dm-3<\/sup> in aqueous solutions.
\n\u2022 Therefore it is convenient to express the [H3<\/sub>O+<\/sup>] of solutions on logarithmic scale. The negative logarithm of hydronium ions or hydrogen ions concentration is called pH. The ‘p’ in the term pH means power and ‘H+<\/sup>‘ stands for hydronium ions or hydrogen ions. The pH is defined mathematically as,\"negative
\n\u2022 The negative sign in the definition means that the pH decreases as the hydrogen ion (hydronium ion) concentration increases. A change of one unit in the pH scale corresponds to tenfold change in the hydrogen ion concentration.<\/p>\n
 <\/p>\n
Calculating the pH of solutions of strong acids<\/h4>\n
\u2022 It is assumed that all strong acids are completely ionised in aqueous solution. The hydronium ion concentration is obtained directly from the concentration of the acid. For monoprotic strong acids [H3<\/sub>O+<\/sup>] is equal to its molar concentration.
\nEg. If the concentration of hydrochloric acid solution is 0.1 mol dm-3<\/sup>.
\nHCl(aq) + H2<\/sub>O(l) \u2192\u00a0H3<\/sub>O+(aq) + Cl–<\/sup>(aq)
\n\u2234 [H3<\/sub>O+(aq) ] = 0.1 mol dm-3<\/sup>
\n-log10<\/sub> 0.1 = 1
\n\u2234 pH = 1<\/p>\n
\u2022 Molecules of diprotic strong acids such as sulphuric acid dissociates completely in dilute solutions.
\nH2<\/sub>SO4<\/sub>(aq) + 2H2<\/sub>O(l) \u2192\u00a02H3<\/sub>O+<\/sup> (aq) + SO4<\/sub>2-<\/sup> (aq)
\nIn 0.1 mol dm-3<\/sup> sulphuric acid solution, [H3<\/sub>O+<\/sup>(aq)] = 0.2 mol dm-3<\/sup> (assuming complete dissociation)
\nlog10<\/sub> 0.2 = -0.7
\n\u2234 -log10<\/sub> 0.2 = 0.7
\n\u2234 pH = 0.7<\/p>\n
<\/h4>\n
Calculating the pH of solutions of strong bases<\/h4>\n
\u2022 Strong bases are also assumed to be completely ionised. The hydroxide ion concentration is therefore easily obtained from the concentration of the base.
\neg. Consider 0.3 mol dm-3<\/sup> solution of sodium hydroxide.
\nNaOH (aq) \u2192\u00a0Na+<\/sup>(aq) + OH–<\/sup>(aq)
\n[OH–<\/sup>(aq)] = 0.3 mol dm-3<\/sup><\/p>\n
\u2022 The hydrogen ion concentration of the solution is determined by the ionic product of water. At 298 K the ionic product is 1×10-14 mol2<\/sup> dm-6<\/sup>. Hydrogen ion concentration at this temperature is given by<\/p>\n
\"\"<\/a><\/p>\n
 <\/p>\n
 <\/p>\n
Calculating the pH of solutions of weak acids<\/h4>\n
\u2022 A weak acid does not ionise fully in aqueous solution<\/p>\n
\"\"<\/p>\n
By applying equilibrium law,<\/span><\/p>\n
\"\"<\/span><\/p>\n
the degree of dissociation is \u03b1 and the initial concentration of HA is C<\/span><\/p>\n
\"\"<\/span><\/p>\n
\"\"<\/span><\/p>\n
\"deriavtion<\/p>\n
pH<\/sup> = -log10<\/sub>\u221a(KaC)<\/p>\n
Similarly<\/p>\n
pOH<\/sup> = -log10<\/sub>\u221a(KbC)<\/p>\n
 <\/p>\n
pH of a solution of salt of strong acid and strong base<\/strong><\/span><\/h4>\n
Let’s take the simplest example absolute;<\/span> sodium chloride, NaCl.<\/strong><\/span> The constituent ions are Na<\/strong>+<\/strong><\/sup> and Cl<\/strong>–<\/strong><\/sup>.\u00a0<\/strong><\/span>The acid-base theory of Bronsted-Lowry has taught us that the more strong the base the\u00a0weaker it will be its conjugate acid and vice versa, the\u00a0more strong the acid the weaker will be its conjugate base.<\/span><\/p>\n
In this case, we have Na+<\/sup><\/strong>, which is the conjugate acid of NaOH<\/strong>, and Cl<\/strong>–<\/strong><\/sup>, which is the conjugate base of HCl.<\/strong><\/span> NaOH is a strong base, and in the same way HCl is a strong acid.<\/span> It follows that the Na+<\/sup> acid behavior and Cl–<\/sup> <\/strong>basic behavior are totally irrelevant.<\/span><\/p>\n
If none of the two ions interacts with H2<\/sub>O<\/strong>\u00a0then we will have no pH changes.<\/span><\/p>\n
It follows that the pH of an aqueous solution of NaCl is neutral, 7.<\/span><\/p>\n
The salts derived from strong acid and strong base leave unchanged the pH of a solution.<\/p>\n
 <\/p>\n
pH of a solution of a salt formed from a weak acid and a strong base<\/strong><\/span><\/h4>\n
Take for example the sodium acetate, CH3<\/sub>COONa.<\/strong><\/span> And ‘it is including respectively CH3<\/sub>COO<\/strong>–<\/strong><\/sup> and Na<\/strong>+<\/strong><\/sup>.<\/strong><\/span> Na+<\/sup> is the conjugate acid of NaOH,<\/strong>and as we saw in the previous example, its contribution to the pH is null.<\/span> CH3<\/sub>COO–<\/sup> instead it is the conjugate base of CH3<\/sub>COO<\/strong>H,<\/strong> a weak acid.<\/span> As the conjugate base of\u00a0a weak acid, it shows a certain affinity towards the proton, and it behaves then as a base:<\/span><\/p>\n
\"\"<\/p>\n
The acid-base reaction of CH3<\/sub>COO–<\/sup> with water is called “hydrolysis.”<\/span> In this case we are speaking of basic hydrolysis, because the reaction generates an excess of\u00a0OH–\u00a0<\/sup><\/strong>.<\/span>Since the reaction generates this excess, the salt of a weak acid and a strong base in water give a pH> 7, \u200b\u200btherefore alkaline.<\/strong><\/span> For the calculation of pH, we can consider the solution of this salt exactly as the solution of just a weak base (in our case CH3<\/sub>COO–<\/sup>).<\/span> We have seen that the formula for calculating<\/span> \"\" generated resulting from a weak base is:<\/span><\/p>\n
\"\"<\/p>\n
The K<\/strong>b<\/strong><\/sub>\u00a0of the weak base\u00a0CH3<\/sub>COO<\/strong>–<\/strong><\/sup> can be obtained from Ka<\/sub> of<\/strong>\u00a0acetic acid as it follows:<\/span><\/p>\n
\"\"\"\"<\/p>\n
The base concentration<\/span> \"\" in this case is the concentration<\/span> \"\" of our salt.<\/span><\/p>\n
From this considerations\u00a0the formula for calculating the<\/span> \"\" a salt-strong base weak acid solution:<\/span><\/p>\n
\"\"<\/p>\n
pOH = -log10<\/sub>\u221a(KwC\/Ka)\u00a0<\/span><\/p>\n
 <\/p>\n
pH of solution of \u00a0salt formed from weak base and strong acid<\/strong><\/span><\/h4>\n
Take for example the ammonium chloride, NH4<\/sub>Cl.<\/strong><\/span> Its constituent ions are NH4<\/sub><\/strong>+<\/strong><\/sup> and Cl<\/strong>–\u00a0<\/strong><\/sup>.<\/span>NH4<\/sub>+<\/sup> is the conjugate acid of NH3<\/sub><\/strong>,<\/sub> ammonia, a weak base.<\/span> It will therefore have a tendency, altough small, to cede that acid hydrogen.<\/span> Cl–<\/sup> instead is the conjugate base of HCl<\/strong>, and as we have seen in the examples above its contribution to the pH is practically zero, and then we can overlook it.<\/span> We can\u00a0simply consider a solution of this salt as an aqueous solution of the weak acid NH4<\/sub>+<\/sup> (which will then give an acid pH)<\/strong><\/span><\/p>\n
\"\"<\/p>\n
The approximate formula to calculate the pH of a weak acid is the following:<\/span><\/p>\n
\"\"<\/p>\n
The Ka<\/sub><\/strong> of the ammonium ion can be easily obtained from the Kb<\/sub><\/strong>\u00a0of ammonia.<\/span> The concentration of acid in the formula, Ca <\/strong>, is therefore in this case the concentration Cs<\/sub> <\/strong>of our salt.<\/span><\/p>\n
\"\"<\/p>\n
<\/h4>\n
pH of a solution of salt formed from weak acid and weak base<\/strong><\/span><\/h4>\n
Let’s take for example ammonium acetate, CH3<\/sub>COONH<\/strong>4.<\/strong><\/sub><\/span> The constituent ions are\u00a0CH3<\/sub>COO<\/strong>–<\/strong><\/sup> and NH4<\/sub><\/strong>+.<\/strong><\/sup><\/span> Both of these ions are able to give hydrolysis (as we have seen in the previous examples); CH3<\/sub>COO<\/strong>–<\/strong><\/sup> gives basic hydrolysis, while acid hydrolysis is given by\u00a0NH4<\/sub>+\u00a0<\/sup><\/strong>.<\/span>The pH of a salt composed of weak acid and weak base will depend on the strength relationship of the acidic and basic components of the salt.\u00a0<\/span>If the base strength and the acid strength are equal the result is a neutral solution.<\/span> And that’s right the case for example of CH3<\/sub>COONH4<\/sub><\/strong><\/span> .<\/span><\/p>\n
Ka <\/sub>NH4<\/sub>+\u00a0<\/sup>\u00a0<\/sub>= \u00a0Kb<\/sub> CH3<\/sub>COO–<\/sup><\/strong><\/span><\/p>\n
If base strength is prevailing on acid strength the salt\u00a0will give the solution an alkaline pH.<\/span> And that’s the case for example of NH4<\/sub>CN.<\/strong><\/span><\/p>\n
Kb <\/sub>CN– \u00a0<\/sup>> Ka<\/sub>\u00a0NH4<\/sub>+<\/sup><\/strong><\/span><\/p>\n
If the prevailing acid strength on the base strength of course the result will be an acid pH.<\/span><\/p>\n
In any case if we want to exactly calculate the pH of a salt formed by weak acid and weak base (also what we called a neutral solution if we wanted to be accurate, has to be calculated like this because Ka<\/sub> and Kb<\/sub> are always at least slightly different), there is a very simple formula:<\/span><\/p>\n
\"\"<\/p>\n
In which:<\/span><\/p>\n
– Ka<\/sub><\/strong> is the constant of the conjugate acid of the weak base which forms the salt.<\/span> For example, in the case of CH3<\/sub>COONH<\/strong>4<\/strong><\/sub> would be the Ka<\/sub> of CH3<\/sub>COOH.<\/strong><\/span><\/p>\n
– K<\/strong>b<\/strong><\/sub> is the constant of the conjugate base of the weak acid that constitutes the salt.<\/span> For example, in the case of CH3<\/sub>COONH4<\/sub><\/strong> would be the Kb<\/sub>of NH3.<\/sub><\/strong><\/span><\/p>\n
 <\/p>\n
pH Determination<\/strong><\/p>\n
The pH value of a solution can be determined using
\n\u2022 indicators
\n\u2022 indicator papers
\n\u2022 pH meter<\/p>\n
 <\/p>\n
Theory of indicators<\/p>\n
Any acid – base indicator is a weak acid or a weak base. They show one colour below a certain pH value and another colour above a certain pH value.
\nConsider acid-base indicator as weak acid which can be represented as HIn which takes part in the following equilibrium.
\nHIn(aq) + H2<\/sub>O(l) \u00a0 \u21cc \u00a0<\/span><\/span><\/span><\/span><\/span><\/b><\/strong>H3<\/sub>O+<\/sup>(aq) + In–<\/sup>(aq)
\nColour I \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Colour II<\/p>\n
Instead of Ka the equilibrium is described by the constant KIn to show that indicator is used.<\/p>\n
KIn = [H3<\/sub>O+<\/sup>(aq)][In–<\/sup>(aq)]\/[HIn(aq)]<\/p>\n
-log10<\/sub>KIn = -log10<\/sub>[H3<\/sub>O+<\/sup>(aq)]- -log10<\/sub>[In–<\/sup>(aq)]- (-log10<\/sub>[HIn(aq)])<\/p>\n
pKIn = pH + log10<\/sub>[In–<\/sup>(aq)]\/[HIn(aq)]<\/p>\n
 <\/p>\n
The HIn\u00a0and In–<\/sup>\u00a0are of two different colours.<\/p>\n
The addition of acid to this solution will push the equilibrium to the left and the colour of HIn will become prominent.
\nAddition of alkali will result in the equilibrium moving to the right since the OH –<\/sup>\u00a0 will remove the H3<\/sub>O+<\/sup>\u00a0ion from the equilibrium and the colour of In–<\/sup>\u00a0will become prominent.<\/p>\n
Colour I displayed at a lower pH may be called ‘lower colour’ and the colour II shown at higher pH can be called the’ higher colour’.<\/p>\n
When the pH of the solution becomes equal to pKIn, an intermediate colour of an equimolar mixture of HIn and In–<\/sup>\u00a0\u00a0is shown.<\/p>\n
When the pH range of a particular indicator is equal to pKIn + 1 and pKIn – 1 the colour change occurs over a pH range of about 2 units.<\/p>\n
 <\/p>\n
\u2022 To select the suitable indicator for a particular titration, pH range of the colour change of the indicator should be in the abrupt pH range (vertical portion) that embraces equivalence point of the titration.
\n\u2022 If pH at equilibrium of a titration is equal to the pKIn of an indicator, that indicator fits well for that titration.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n
Indicator<\/b><\/td>\nColor<\/b><\/td>\n<\/td>\npKIn<\/sub><\/b><\/td>\npH range<\/b><\/td>\n<\/tr>\n
<\/td>\nAcid<\/u><\/td>\nBase<\/u><\/td>\n<\/td>\n<\/td>\n<\/tr>\n
Thymol Blue<\/a> – 1st<\/sup> change<\/td>\nred<\/td>\nyellow<\/td>\n1.5<\/td>\n1.2 – 2.8<\/td>\n<\/tr>\n
Methyl Orange<\/a><\/td>\nred<\/td>\nyellow<\/td>\n3.7<\/td>\n3.2 – 4.4<\/td>\n<\/tr>\n
Bromocresol Green<\/a><\/td>\nyellow<\/td>\nblue<\/td>\n4.7<\/td>\n3.8 – 5.4<\/td>\n<\/tr>\n
Methyl Red<\/a><\/td>\nyellow<\/td>\nred<\/td>\n5.1<\/td>\n4.8 – 6.0<\/td>\n<\/tr>\n
Bromothymol Blue<\/a><\/td>\nyellow<\/td>\nblue<\/td>\n7.0<\/td>\n6.0 – 7.6<\/td>\n<\/tr>\n
Phenol Red<\/a><\/td>\nyellow<\/td>\nred<\/td>\n7.9<\/td>\n6.8 – 8.4<\/td>\n<\/tr>\n
Thymol Blue<\/a> – 2nd<\/sup> change<\/td>\nyellow<\/td>\nblue<\/td>\n8.9<\/td>\n8.0 – 9.6<\/td>\n<\/tr>\n
Phenolphthalein<\/a><\/td>\ncolorless<\/td>\nmagenta<\/td>\n9.4<\/td>\n8.2 – 10.0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
 <\/p>\n
Calculation of pH value of acid-base reactions at different points and drawing pH curves<\/p>\n
\u2022 The point at which the reaction completed is called the equivalence point. At the equivalence point of a strong acid- strong base titration,
\n[H +<\/sup>(aq)] = [ OH –<\/sup>\u00a0(aq)]<\/p>\n
\u2022 In strong acid \u2013strong base titration neither cation (eg. Na+<\/sup>, K+<\/sup>) nor the anion (eg. HCl–<\/sup>,NO3<\/sub>–<\/sup>) undergoes hydrolysis at the equivalence point. Therefore, pH at equivalence point is 7.0. Here pH at the equivalence point is determined only by the dissociation of H2<\/sub>O.<\/p>\n
\u2022 But in other titrations cations of weak base and anions of weak acid undergo hydrolysis and determine the pH of the resulting solution.<\/p>\n
 <\/p>\n
Calculation of pH value of acid-base reactions at different points and drawing pH curves<\/h3>\n
    \n
  • The term “equivalence point” means that the solutions have been mixed in exactly the right proportions according to the equation.<\/li>\n
  • The term “end point” is where the indicator changes colour. As you will see on the page about indicators, that isn’t necessarily exactly the same as the equivalence point.<\/li>\n<\/ul>\n
     <\/p>\n
    Titration curves<\/h3>\n
    1) Strong acid\/ strong base titration<\/h4>\n
    We’ll take hydrochloric acid and sodium hydroxide as typical of a strong acid and a strong base.<\/p>\n
    <\/p>\n
    <\/p>\n
    \u2022 Titration shows a rapid change in pH from 3 – 11 around its equivalence point. Hence any indicator having pKIn in this range can be used to detect the end point of the titration.
    \neg. Methyl orange (pKIn = 3.7), Bromothymol blue (pKIn = 7),Phenolpthalene (pKIn = 9.6)
    \n\u2022 When concentrations of reactants are low the range in which a rapid change is shown becomes narrow.<\/p>\n
    <\/p>\n
     <\/p>\n
    2) Strong base \/ weak acid titration<\/h4>\n
    We’ll take ethanoic acid and sodium hydroxide as typical of a weak acid and a strong base.<\/p>\n
    <\/p>\n
    <\/p>\n
    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.<\/p>\n
    For the above titration a rapid change in pH shows around 7.5 to 10. Therefore, the suitable indicator is phenolpthalene. (pKIn = 9.6)<\/p>\n
     <\/p>\n
    3. Strong acid\/ weak base titration<\/h4>\n
    <\/p>\n
    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.<\/p>\n
    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.<\/p>\n
    For the above titration a rapid change in pH shows around 3 to 6. Therefore, one of the
    \nsuitable indicators is methyl orange.<\/p>\n
     <\/p>\n
    (4) For weak acid – weak base titrations there is no rapid change in pH around the equivalence point. Therefore, it is very difficult to accurately detect the end point by using an indicator.<\/p>\n
     <\/p>\n
    Adding hydrochloric acid to sodium carbonate solution<\/b><\/h4>\n
    The overall equation for the reaction between sodium carbonate solution and dilute hydrochloric acid is:<\/p>\n
    <\/p>\n
    If you had the two solutions of the same concentration, you would have to use twice the volume of hydrochloric acid to reach the equivalence point – because of the 1 : 2 ratio in the equation.<\/p>\n
    Suppose you start with 25 cm3<\/sup> of sodium carbonate solution, and that both solutions have the same concentration of 1 mol dm-3<\/sup>. That means that you would expect<\/i> the steep drop in the titration curve to come after you had added 50 cm3<\/sup> of acid.<\/p>\n
    The actual graph looks like this:<\/p>\n
    <\/p>\n
    The graph is more complicated than you might think – and curious things happen during the titration.<\/p>\n
    You expect carbonates to produce carbon dioxide when you add acids to them, but in the early stages of this titration, no carbon dioxide is given off at all.<\/p>\n
    Then – as soon as you get past the half-way point in the titration – lots of carbon dioxide is suddenly released.<\/p>\n
    The graph is showing two end points – one at a pH of 8.3 (little more than a point of inflexion), and a second at about pH 3.7. The reaction is obviously happening in two distinct parts.<\/p>\n
    In the first part, complete at A<\/b> in the diagram, the sodium carbonate is reacting with the acid to produce sodium hydrogencarbonate:<\/p>\n
    <\/p>\n
    You can see that the reaction doesn’t produce any carbon dioxide.<\/p>\n
    In the second part, the sodium hydrogencarbonate produced goes on to react with more acid – giving off lots of CO2<\/sub>.<\/p>\n
    <\/p>\n
    That reaction is finished at B<\/b> on the graph.<\/p>\n
    It is possible to pick up both of these end points by careful choice of indicator. That is explained on the separate page on indicators.<\/p>\n","protected":false},"excerpt":{"rendered":"
    Ionisation of water and Ionic Product of Water ( Kw) Pure water being a weak electrolyte under goes self ionization to a small extent as follows: The equilibrium constant for this reaction is: The concentration of unionized water is taken as constant because the degree on ionization of water is very small. 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