Q.10.100 - An Introduction to General, Organic, and Biological Chemistry

Question

A buffer solution is made by dissolving ${\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2 \mathrm{and} {\mathrm{NaC}}_2{\mathrm{H}}_3{\mathrm{O}}_2$ in water.

a. Write an equation that shows how this buffer neutralizes added acid.

b. Write an equation that shows how this buffer neutralizes added base.

Step-by-Step Solution

Answer

Part (a) . ${\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}\to {\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{H}}_2\mathrm{O}$

Part (b) . ${\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{OH}}^{-}\to 2{\mathrm{H}}_2\mathrm{O}$

Step 1: Given Information (Part a)

A buffer solution is a solution which resists any adjustment of pH esteem in any event when a modest quantity of acid or base is added to it. there are two types of buffer: Acid Buffer which is a mixture of frail acid and salt of this weak base area of strength. Basic buffer is the mixture of a weak base and salt of this weak base with strong acid.

Step 2: Explanation Part (a)

An acidic buffer is made by dissolving ${\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2\text{ and }{\mathrm{NaC}}_2{\mathrm{H}}_3{\mathrm{O}}_2$ in water.

The reaction with water is,

$\begin{array}{r}{\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}\\ {\mathrm{CH}}_3\mathrm{COONa}\to {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{Na}}^{+}\end{array}$

The equation shows how this buffer neutralizes added acid,

${\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}\rightleftarrows {\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{H}}_2\mathrm{O}$

Step 3: Explanation Part (b)

An acidic buffer is made by dissolving ${\mathrm{HC}}_2{\mathrm{H}}_3{\mathrm{O}}_2\text{ and }{\mathrm{NaC}}_2{\mathrm{H}}_3{\mathrm{O}}_2$ in water.

The reaction with water is,

${\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}$

${\mathrm{CH}}_3\mathrm{COONa}\to {\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{Na}}^{+}$

The equation shows how this buffer neutralizes added base,

${\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{OH}}^{-}\to 2{\mathrm{H}}_2\mathrm{O}$