Buffer Solutions

A buffer solution is a solution which resists a change in its pH when such a change is caused by the addition of a small amount of acid or base. This does not mean that the pH of the buffer solution does not change (we make this assumption while doing numerical problems). It only means that the change in pH would be less than the pH that would have changed for a solution that is not a buffer.

There are two types of buffer solutions:

(i) Acidic buffer (ii) Basic buffer

(i) Acidic buffer: It is the solution of a weak acid and it salts with strong base.

e.g. (CH₃COOH + CH₃COONa), (Phthalic acid + Potassium acid phthalate), (Boric acid + Borax)

(ii) Basic buffer: It is the solution of a weak base and its salt with strong acid.

e.g. (NH₄OH + HCl), (Glycine + Glycine hydrochloride) etc.

(iii) Salt buffer: Aqueous solution of the salt of weak acid and weak base e.g. CH₃COONH₄.

Explanation of acidic buffer

Since acidic buffer is the mixture of weak acid and its salt with strong base e.g. CH₃COOH and CH₃COONa.

They have the following equilibria in solution.

$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}$ (Feebly ionised)

$C{H}_{3}COONa\to C{H}_{3}CO{O}^{-}+N{a}^{+}$ (Completely ionised)

$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$ (Very feebly ionised)

The salt is completely ionised. Due to common ion effect CH₃COO^– ion from salt (CH₃COONa) suppresses the ionisation of CH₃COOH. Thus acid buffer is a mixture of more unionised CH₃COOH molecule and more CH₃COO^– ion that obtained from ionisation of salt alone.

Addition of acid: If a drop of HCl is added, H⁺ ion from HCl combine with CH₃COO^– ion to form feebly ionised CH₃COOH molecule.

$HCl\to H^{+}+C{l}^{-}$

$C{H}_{3}CO{O}^{-}+\underset{fromHCl}{H^{+}}\rightleftharpoons C{H}_{3}COOH$

Addition of base

If a drop of NaOH is added, OH^– ion from NaOH combine with CH₃COOH molecule to form feebly ionised H₂O molecule.

$NaOH\to N{a}^{+}+O{H}^{-}$

\begin{align}  C{{H}_{3}}COOH+O{{H}^{-}}\xrightarrow[{}]{{}}C{{H}_{3}}CO{{O}^{-}}+{{H}_{2}}O \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,from\,\,NaOH \\ \end{align}

Thus small addition of H⁺ and OH^– to buffer solution makes no appreciable change in pH of the buffer.

Calculation of pH of acidic buffer

It consist of a weak acid which ionises as

$HA\stackrel{k_a}{\rightleftharpoons }{H}^{+}+A^{-}$

where K_a is ionisation constant.

$K_a=\frac{[H^{+}][A^{-}]}{[HA]}\Rightarrow [H^{+}]=\frac{K_a[HA]}{[A^{-}]}$………… (1)

In preparation of buffer salt highly ionised salt BA is also added which ionise completely.

$BA\to B^{+}+A^{-}$

Due to common ion effect, A^– ions from salt BA suppresses the ionisation of weak acid.

Thus, $[BA]\equiv [A^{-}]$

${[HA]}_{\text{equilibrium}}={[HA]}_{\text{initial}}=[acid]$

and [A^–] at equilibrium = [BA] = [salt]

$[H^{+}]=K_a\frac{[acid]}{[salt]}$

Taking log and reversing the sign.

$-{\log }_{10}[H^{+}]=-{\log }_{10}{K}_a-\log \frac{[acid]}{[salt]}$

$pH=-{\log }_{10}{K}_a+\log \frac{[salt]}{[base]}$

$pH=p{K}_a+\log \frac{[salt]}{[base]}$

This is called Hendersons equation.

Basic Buffers

Having pH towards base side. It is a mixture of weak base and its salt with strong acid.

e.g. NH₄OH + NH₄Cl

Explanation:

$N{H}_{4}OH\rightleftharpoons N{H}_4^{+}+O{H}^{-}$

$N{H}_{4}Cl\to N{H}_4^{+}+C{l}^{-}$

The salt is almost completely ionised. Due to common ion effect $N{H}_4^{+}$ ion from NH₄Cl suppresses the ionisation of NH₄OH. Thus a basic buffer is a mixture of more unionised NH₄OH molecule and more $N{H}_4^{+}$ ions that are obtained from ionisation of NH₄Cl alone.

Addition of acid:

If a drop of HCl is added the H⁺ from HCl combine with NH₄OH to form feebly ionised H₂O molecules.

$HCl\to H^{+}+C{l}^{-}$

\begin{gathered} N{H_4}OH + {H^ + } \rightleftharpoons NH_4^ + + {H_2}O \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,feebly\,\,ionised \hfill \\ \end{gathered}

Addition of base:

If a drop of NaOH is added OH^– ion from NaOH combine with $N{H}_4^{+}$ ion forming feebly ionised NH₄OH molecules.

$NaOH\to N{a}^{+}+O{H}^{-}$

$N{H}_4^{+}+O{H}^{-}\rightleftharpoons N{H}_{4}OH$

Hence small addition of acid or base to basic buffer solution makes no appreciable change in pH.

Calculation of pH of basic buffers

It consist of a weak base which ionises as

$BOH\stackrel{k_b}{\rightleftharpoons }{B}^{+}+O{H}^{-}$, where K_b is ionisation constant of weak base.

$K_b=\frac{[B^{+}][O{H}^{-}]}{[BOH]}\Rightarrow [O{H}^{-}]=\frac{K_b[BOH]}{[B^{+}]}$

In preparation of basic buffers a highly ionised salt BA is used. It ionised as

$BA\to B^{+}+A^{-}$

Due to common ion effect, B⁺ ions from salt BA suppresses the ionisation of BOH. It can be assumed that all B⁺ ions in basic buffers are from salt BA.

i.e. [BA] = [B⁺]

or [BOH] at equilibrium = [base]

${[B^{+}]}_{\text{equilibrium}}=[BA]=[salt]$

$[O{H}^{-}]=K_b\frac{[salt]}{[base]}$

Taking log and reversing the signs,

$-{\log }_{10}[O{H}^{-}]=-{\log }_{10}{K}_b-{\log }_{10}\frac{[base]}{[salt]}$

$pOH=p{K}_b+{\log }_{10}\frac{[salt]}{[base]}$

This is called Hendersons equation.

Illustration 1. CH₃COOH (50 ml, 0.1M) is titrated against 0.1M NaOH solution. Calculate the pH at the addition of 0 ml, 10 ml, 20 ml, 25 ml, 40 ml, 50 ml and 60 ml of NaOH. K_a of CH₃COOH is 2 × 10^–5.

Solution: (i) When 0 ml of NaOH is added, the pH is due to acetic acid,

$\therefore$ [H⁺] = $\sqrt{K_a\times C}$ = $\sqrt{2\times {10}^{-5}\times 0.1}$ = $\sqrt{2\times {10}^{-6}}$

pH = – log$\sqrt{2\times {10}^{-6}}$ = –$\frac{1}{2}$ [log 2–6] = 3–0.15 = 2.85

(ii) When 10 ml of NaOH is added, it reacts with CH₃COOH to produce salt and water. The solution is then a buffer.

$\therefore$pH = pK_a + log $\frac{\left[Conjugate\text{ base}\right]}{\left[Acid\right]}$

= 4.699 + log $\frac{10\times 0.1}{50\times 0.1-10\times 0.1}$ = 4.699 + log $\frac{1}{4}$= 4.0969

(iii) When 20 ml of NaOH is added.

pH = pK_a + log$\frac{20\times 0.1}{50\times 0.1-20\times 0.1}$ = 4.699 + log $\frac{2}{3}$ = 4.5229

(iv) When 25 ml of NaOH is added,

pH = 4.699 + log $\frac{25\times 0.1}{50\times 0.1-25\times 0.1}$ = 4.699

(v) When 40 ml of NaOH is added,

pH = 4.699 + log $\frac{40\times 0.1}{50\times 0.1-40\times 0.1}$= 4.699 + log 4 = 5.3011

(vi) When 50 ml of NaOH is added,

Here, if we use the buffer equation, pH would be = $\infty$

But we can't use the buffer equation as there is no acid. Therefore we used the hydrolysis equation.

$\therefore$ [H⁺] = $\sqrt{\frac{K_W{K}_a}{C}}$

C = $\frac{0.1}{2}$[ $\therefore$ Total Volume is 100 ml and millimoles of salt is 50 x 0.1]

$\Rightarrow$ [H⁺] = $\sqrt{\frac{{10}^{-14}\times 2\times {10}^{-5}\times 2}{0.1}}$ = pH = 8.699(vii) When 60 ml of NaOH is added, the excess OH^– ion from NaOH would suppress the hydrolysis of CH₃COO^– ion. So we can ignore the contribution of OH^– ion from the hydrolysis of CH₃COO^– ion.$\therefore$ [OH^–] = $\frac{0.1\times 10}{110}$ [10 ml of OH^– ion is in excess] = $\frac{1}{110}$

pOH = 2.0414

$\therefore$ pH = 14 – 2.0414 = 11.9586