Q.10.99 - An Introduction to General, Organic, and Biological Chemistry

Question

A buffer solution is made by dissolving ${\mathrm{H}}_3{\mathrm{PO}}_4\text{ and }{\mathrm{NaH}}_2{\mathrm{PO}}_4$ 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) . ${\mathrm{H}}_2{{\mathrm{PO}}_4}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}\to {\mathrm{H}}_3{\mathrm{PO}}_4+{\mathrm{H}}_2\mathrm{O}$

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

Step 1: Information Part (a)

The representation of a chemical reaction in the type of substance is known as a chemical equation. The equation in which the quantity of particles of the relative multitude of atoms is equivalent on the two sides of the equation is known as a fair chemical equation.

Step 2: Explanation Part (a)

By creating a buffer solution by adding ${\mathrm{H}}_3{\mathrm{PO}}_4\text{ and }{\mathrm{NaH}}_2{\mathrm{PO}}_4$ in water, we get,

$\begin{array}{r}{\mathrm{H}}_3{\mathrm{PO}}_4+\mathrm{H}2\mathrm{O}\rightleftarrows {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\\ {\mathrm{NaH}}_2{\mathrm{PO}}_4\to {\mathrm{Na}}^{+}+{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\end{array}$

The equation that shows how this buffer neutralizes added acid is,

${\mathrm{H}}_2{{\mathrm{PO}}_4}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}\to {\mathrm{H}}_3{\mathrm{PO}}_4+{\mathrm{H}}_2\mathrm{O}$

Step 3: Explanation Part (b)

By creating a buffer solution by adding ${\mathrm{H}}_3{\mathrm{PO}}_4\text{ and }{\mathrm{NaH}}_2{\mathrm{PO}}_4$in water, we get,

$\begin{array}{r}{\mathrm{H}}_3{\mathrm{PO}}_4+{\mathrm{H}}_2\mathrm{O}\rightleftarrows {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\\ {\mathrm{NaH}}_2{\mathrm{PO}}_4\to {\mathrm{Na}}^{+}+{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\end{array}$

The equation that shows how this buffer neutralizes added base is,

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