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Hydrolysis of Salts and the pH of their Solutions
Hydrolysis of Salts and the pH of their Solutions
Salts formed by the reactions between acids and bases in definite proportions, undergo ionization in water. The cations/anions formed on ionization of salts either exist as hydrated ions in aqueous solutions or interact with water to reform corresponding acids/bases depending upon the nature of salts. The later process of interaction between water and cations/anions or both of salts is called hydrolysis. The pH of the solution gets affected by this interaction. The cations (e.g.,Na+,K+,Ca2+,Ba2+,etc.)of strong bases and anions (e.g., Cl-, Br-, NO\( _3^ - \), ClO\(_4^ - \) etc.) of strong acids simply get hydrated but do not hydrolyse, and therefore the solutions of salts formed from strong acids and bases are neutral i.e., their pH is 7. However, the other category of salts do undergo hydrolysis.
We now consider the hydrolysis of the salts of the following types :
1. salts of weak acid and strong base e.g., CH3COONa.
2. salts of strong acid and weak base e.g.,NH4Cl, and
3. salts of weak acid and weak base, e.g., CH3COONH4.
In the first case, CH3COONa being a salt of weak acid, CH3COOH and strong base, NaOH gets completely ionised in aqueous solution.
CH3COONa(aq)\(\to\)CH3COO– (aq)+ Na+(aq)
Acetate ion thus formed undergoes hydrolysis in water to give acetic acid and OH– ions CH3COO–(aq)+H2O(1)
CH3COOH(aq)+OH–(aq)Acetic acid being a weak acid (Ka = 1.8 × 10–5) remains mainly unionised in solution.This results in increase of OH– ion concentration in solution making it alkaline.The pH of such a solution is more than 7.
Similarly, NH4Cl formed from weak base, NH4OH and strong acid, HCl, in water dissociates completely.
NH4Cl(aq) \(\to\)NH\(_4^ + \)(aq) +Cl– (aq)
Ammonium ions undergo hydrolysis with water to form NH4OH and H+ ions
NH\(_4^ + \)(aq)+H2O(1)
NH4OH(aq)+H+(aq)Ammonium hydroxide is a weak base (Kb= 1.77 × 10–5) and therefore remains almost unionised in solution.This results in increased of H+ ion concentration in solution making the solution acidic. Thus, the pH of NH4Cl solution in water is less than 7.
Consider the hydrolysis of CH3COONH4 salt formed from weak acid and weak base.The ions formed undergo hydrolysis as follow:
CH3COO– + NH\(_4^ + \)+ HO
CH3COOH +NH4OHCH3COOH and NH4OH, also remain into partially dissociated form:
CH3COOH
CH3COO– + H+NH4OH
NH \(_4^ + \)+ OH–H2O
H++OH-Without going into detailed calculation, it can be said that degree of hydrolysis is independent of concentration of solution, and pH of such solutions is determined by their pK values:
pH = 7 + ½ (pKa – pKb) (7.38)
The pH of solution can be greater than 7, if the difference is positive and it will be less than 7, if the difference is negative.
PROBLEM 25
The pKa of acetic acid and pKb of ammonium hydroxide are 4.76 and 4.75 respectively.Calculate the pH of ammonium acetate solution.
SOLUTION
pH = 7 + ½ [pKa – pKb]
= 7 + ½ [4.76 – 4.75]
= 7 + ½ [0.01] = 7 + 0.005 = 7.005
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Buffer Solutions
BUFFER SOLUTIONS
Many body fluids e.g., blood or urine have definite pH and any deviation in their pH indicates malfunctioning of the body. The control of pH is also very important in many chemical and biochemical processes. Many medical and cosmetic formulations require that these be kept and administered at a particular pH. The solutions which resist change in pH on dilution or with the addition of small amounts of acid or alkali are called Buffer Solutions. Buffer solutions of known pH can be prepared from the knowledge of pKa of the acid or pKb of base and by controlling the ratio of the salt and acid or salt and base. A mixture of acetic acid and sodium acetate acts as buffer solution around pH 4.75 and a mixture of ammonium chloride and ammonium hydroxide acts as a buffer around pH 9.25. You will learn more about buffer solutions in higher classes.
Designing Buffer Solution
Knowledge of pKa, pKb and equilibrium constant help us to prepare the buffer solution of known pH. Let us see how we can do this.
Preparation of Acidic Buffer
To prepare a buffer of acidic pH we use weak acid and its salt formed with strong base. We develop the equation relating the pH, the
equilibrium constant, Ka of weak acid and ratio of concentration of weak acid and its conjugate base. For the general case where the weak acid
HA ionises in water,
HA+H2O
H3O++A-For which we can write the expression
\({K_a} = \frac{{\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {{A^ - }} \right]}}\)
Rearranging the expression we have,
\(\left[ {{H_3}{O^ + }} \right] = {K_a}\frac{{\left[ {HA} \right]}}{{\left[ {{A^ - }} \right]}}\)
Taking logarithm on both the sides and rearranging the terms we get —
\(\begin{array}{l} p{K_a} = pH - \log \frac{{\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Or\\ pH = p{K_a} + \log \frac{{\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}\,\,\,\,\left( {7.39} \right)\\ pH = p{K_a} + \log \frac{{\left[ {Conjugate\,\,\,base,{A^ - }} \right]}}{{\left[ {Acid,HA} \right]}}\,\,\,\,\,\left( {7.40} \right) \end{array}\)
The expression (7.40) is known as Henderson–Hasselbalch equation.The quantity \( \frac{{\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}\) is the ratio of concentration of conjugate base (anion) of the acid and the acid present in the mixture.Since acid is a weak acid, it ionises to a very little extent and concentration of [HA] is negligibly different from concentration of acid taken to form buffer. Also, most of the conjugate base, [A—], comes from the ionisation of salt of the acid.Therefore, the concentration of conjugate base will be negligibly different from the concentration of salt. Thus, equation (7.40) takes the form:
\( pH = p{K_a} + \log \frac{{\left[ {Salt} \right]}}{{\left[ {Acid} \right]}}\)
In the equation (7.39), if the concentration of [A—] is equal to the concentration of [HA], then pH = pKa because value of log 1 is zero.Thus if we take molar concentration of acid and salt (conjugate base) same,the pH of the buffer solution will be equal to the pKa of the acid.So for preparing the buffer solution of the required pH we select that acid whose pKa is close to the required pH. For acetic acid pKa value is 4.76, therefore pH of the buffer solution formed by acetic acid and sodium acetate taken in equal molar concentration will be around 4.76.
A similar analysis of a buffer made with a weak base and its conjugate acid leads to the result,
\( pOH = p{K_b} + \log \frac{{\left[ {Conjugate\,\,acid,B{H^ + }} \right]}}{{\left[ {Base,B} \right]}}\) (7.41)
pH of the buffer solution can be calculated by using the equation pH + pOH =14.
We know that pH + pOH = pKw and pKa + pKb = pKw. On putting these values in equation (7.41) it takes the form as follows:
\(\begin{array}{l} p{K_w} - pH = p{K_w} - p{K_a} + \log \frac{{\left[ {Conjugate\,\,acid,B{H^ + }} \right]}}{{\left[ {Base,B} \right]}}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Or\\ pH = p{K_a} + \log \frac{{\left[ {Conjugate\,\,acid,B{H^ + }} \right]}}{{\left[ {Base,B} \right]}}\,\,\,\,\,\,\,\,\left( {7.42} \right) \end{array}\)
If molar concentration of base and its conjugate acid (cation) is same then pH of the buffer solution will be same as pKa for the base. pKa value for ammonia is 9.25; therefore a buffer of pH close to 9.25 can be obtained by taking ammonia solution and ammonium chloride solution of same molar concentration. For a buffer solution formed by ammonium chloride and ammonium hydroxide, equation (7.42) becomes:
\(pH = 9.25 + \log \frac{{\left[ {Conjugate\,\,\,acid,B{H^ + }} \right]}}{{\left[ {Base,B} \right]}}\)
pH of the buffer solution is not affected by dilution because ratio under the logarithmic term remains unchanged