Chapter 16, Problem 16.60QP

Interpretation Introduction

Interpretation:

The original molarity of the given solution whose ${\text{K}}_a\text{is}3.5\times {10}^{-5}$ and pH is 5.26 at ${25}^{\circ }\text{C}$ has to be calculated

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

Acids ionize in water.  Strong acids ionize completely whereas weak acids ionize to some limited extent.

The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization.

The ionization of a weak acid $\text{HA}$ can be given as follows,

${\text{HA}}_{\text{(aq)}}\to {\text{ H}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

$K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where,

$K_a$ is acid ionization constant,

$[{\text{H}}^{\text{+}}\text{]}$ is concentration of hydrogen ion

$[{\text{A}}^{\text{-}}\text{]}$ is concentration of acid anion

$[\text{HA]}$ is concentration of the acid

$\text{pH}$ definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

$\text{pH}\text{=}{\text{-log[H}}^{\text{+}}\text{]}$

To Calculate: The molarity of the given acid solution whose ${\text{K}}_a\text{is}3.5\times {10}^{-5}$ and  pH is 5.26  at ${25}^{\circ }\text{C}$