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/.