Question

How do you solve chemistry buffer problems?

Answer 1

Hint: To determine the answer of this question we should know what buffer is, what Henderson equation is. The solutions which resist the change in pH are known as buffer solutions. Buffer solution contains weak acid/ base and their salts. We use the relation in pH/pOH, dissociation constant and concentration of acid/ base and their salts to determine the pH of buffer solution.

Complete step by step answer:
Buffer definition: An aqueous solution of weak acid and its conjugate base or weak base and its conjugate acid is known as buffer.

Buffering principle: The acid and conjugate base remains in equilibrium.
$\mathop {{\text{HA}}\,}\limits_{{\text{acid}}} \to \,{{\text{H}}^{\text{ + }}} + \,\,\mathop {{{\text{A}}^ - }}\limits_{{\text{conjugate}}\,\,{\text{base}}} $
- When we add strong acid means simply we are adding hydrogen ions. These added hydrogen ions are utilised by conjugate base to form the acid so, in solution, the hydrogen ion concentration remains the same so pH does not change.
$ \Rightarrow \mathop {{{\text{A}}^ - }}\limits_{{\text{conjugate}}\,\,{\text{base}}} + \,{{\text{H}}^{\text{ + }}} \to \,\,\,\mathop {{\text{HA}}\,}\limits_{{\text{acid}}} $
- Similarly, when we add a strong base, simply we are adding hydroxide ions. These added hydroxide ions are utilized by acid to form conjugate base so, in solution the hydroxide ion concentration remains the same so pH does not change.
$ \Rightarrow \mathop {{\text{HA}}\,}\limits_{{\text{acid}}} + \,{\text{O}}{{\text{H}}^ - } \to \,\,\,\mathop {{{\text{A}}^ - }}\limits_{{\text{conjugate}}\,\,{\text{base}}} + \,{{\text{H}}_{\text{2}}}{\text{O}}$

Henderson Hassel Blech gave an equation to determine the pH of acidic or basic buffer. Which are as follows:
1. For weak acidic buffer:
$ \Rightarrow {\text{pH}}\,\,{\text{ = }}\,{\text{p}}{{\text{K}}_{\text{a}}}\,{\text{ + l}}\,{\text{og}}\,\left[ {\dfrac{{{\text{Base}}}}{{{\text{acid}}}}} \right]$
Where, ${\text{p}}{{\text{K}}_{\text{a}}}$ is the antilog of acid dissociation constant.

2. For weak basic buffer:
$ \Rightarrow {\text{pOH}}\,\,{\text{ = }}\,{\text{p}}{{\text{K}}_{\text{b}}}\,{\text{ + l}}\,{\text{og}}\,\left[ {\dfrac{{{\text{acid}}}}{{{\text{base}}}}} \right]$
Where, ${\text{p}}{{\text{K}}_{\text{b}}}$ is the antilog of base dissociation constant.
For, example, A solution contains $0.2$M of $ {\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$$\left( {{{\text{K}}_{\text{a}}}{\text{ = }}1.2 \times {{10}^{ - 5}}} \right)$ and $0.5$M of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }$. The pH of buffer solution is:
Here, acetic acid is a weak acid which remains in equilibrium with its conjugate base as follows:
$ \Rightarrow {\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}} \to \,{\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }\, + \,{{\text{H}}^ + }$
The Henderson Hassel Blech equation for acidic buffer is,
$ \Rightarrow {\text{pH}}\,\,{\text{ = }}\,{\text{p}}{{\text{K}}_{\text{a}}}\,{\text{ + l}}\,{\text{og}}\,\left[ {\dfrac{{{\text{Base}}}}{{{\text{acid}}}}} \right]$
$ \Rightarrow {\text{pH}}\,\,{\text{ = }}\, - \log (1.2 \times {10^{ - 5}})\,{\text{ + l}}\,{\text{og}}\,\left[ {\dfrac{{0.5}}{{{\text{0}}{\text{.2}}}}} \right]$
$ \Rightarrow {\text{pH}}\,\,{\text{ = }}\, - \log (1.2 \times {10^{ - 5}})\,{\text{ + }}\,0.40$
$ \Rightarrow {\text{pH}}\,\,{\text{ = }}\,4.9\,{\text{ + }}\,0.40$
$ \Rightarrow {\text{pH}}\,\,{\text{ = }}\,5.3$
So, the pH of the acetic acid buffer is $5.3$.

Note: The pH depends upon the hydrogen and hydroxide ion concentrations. pH increases on increasing the concentration of hydroxide ions. The pH decreases on increasing hydrogen ion concentration. The effect of addition of strong acid or base in the buffer is explained by the Le-chatelier principle. The buffer solution is a pH resistance solution. The pH of buffer solution does not alter on the addition of strong acid or base.