51E - Organic Chemistry (Mcmurry)

Question

What is the pH of a 0.050 M solution of formic acid, $\text{p}{K}_a$= 3.75?

Step-by-Step Solution

Answer

The pH of a 0.050 M solution of formic acid is 2.52.

pH value

The term pH is an abbreviation for “power of hydrogen,” where p stands for the German word for “power” and H is the element symbol for hydrogen. The logarithm inverse of H⁺ concentration is pH. The pH value of a system is a type of scale that shows acidity or basicity. When a pH value is high, it is referred to as a basic system. When a system’s pH value is low, it is considered acidic. In order to indicate that there is no acidity or basicity present, the pH value of 7 is considered a neutral pH.

For example, to calculate the pH of an aqueous solution, you need to know the concentration of hydronium ion in moles per liter. So, the pH can be calculated as:

$pH=-\log \left[H_3{O}^{+}\right]$ -----------equation (1)           

p K a value

The exact strength of a given acid HA in water solution is described using the acidity constant ( ) for the acid-dissociation equilibrium.

Example:

 $\begin{array}{l}{\text{HA + H}}_2\text{O }\rightleftarrows {\text{ A}}^{-}{\text{ + H}}_{\text{3}}{\text{O}}^{-}\\ K_a\underset{}{}\text{=}\underset{}{}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{+}{\text{][A}}^{-}\text{]}}{\left[HA\right]}\end{array}$

 

 

Acid strengths are normally using  values rather than  values, where the  is the negative common logarithm of :

$\text{p}K\text{a = -log }{K}_a$ ------------equation (2)

Calculate the pH of formic acid

The  $\text{p}K\text{a }$value of formic acid (HCOOH) is 3.75

So, we can calculate the acid dissociation constant from equation (2)

$\begin{array}{l}p{K}_a=-\log K_a\\ -p{K}_a=\log K_a\\ K_a={10}^{-3.75}\\ Hence\\ =1.8\times {10}^{-4}\end{array}$                                                                              

  

The acid dissociation constant of formic acid is $1.8\times {10}^{-4}$ .

 -$\begin{array}{l}Formicacid\left(HCOOH\right)\underset{}{}+\underset{}{}{H}_{2}O\underset{}{}\stackrel{K_a}{\rightleftarrows }\underset{}{}Formalat{e}^{-}\left(HCO{O}^{-}\right)\underset{}{}+\underset{}{}{H}_3{O}^{+}\\ \left[0.050M\right]\left[x\right]\left[x\right]\end{array}$--------equation (3)

The concentration of formic acid (HCOOH) is 0.050 M.

Let x represent the concentration of H⁺ that dissociates from HCOOH (formic acid), then [HCOOH] = 0.050 – x.

 $K_a\underset{}{}\text{=}\underset{}{}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{+}{\text{][HCOO}}^{-}\text{]}}{\left[HCOOH\right]}$

$K_a=1.8\times {10}^{-4}=\frac{x^2}{0.050-x}$

 

Let 0.050-x = 0.050

 $\begin{array}{l}{x}^2=1.8\times {10}^{-4}\times 0.050\\ x^2=9\times {10}^{-6}\\ x=3\times {10}^{-3}\end{array}$

From equation (2), we have

 $pH=-log\left[H_3{O}^{+}\right]$

From equation (3),  

So,

$\begin{array}{l}pH=-\log \left[x\right]\\ pH=-\log \left[3.0\times {10}^{-3}\right]\\ pH=-\left(-2.52\right)\\ Thus\\ pH=2.52\end{array}$

 

The pH of a 0.050 M formic acid solution is 2.52.