Question

A solution prepared by dissolving 0.100 mole of propionic acid in enough water to make 1.00 L of solution is observed to have a pH = 2.94. What is the Ka for propionic acid?(A) 1.3 x 10¯6(B) 1.1 x 10¯3(C) 1.3 x 10¯5(D) 1.1 x 10¯2

Answer 1

Answer:

C) $k_a=1.3\times 10^{-5}$

Explanation:

pH is defined as the negative logarithm of the concentration of hydrogen ions.

Thus,  

pH = - log [H⁺]

The expression of the pH of the calculation of weak acid is:-

$pH=-log(\sqrt{k_a\times C})$

Where, C is the concentration = 0.5 M

Given, pH = 2.94

Moles = 0.100 moles

Volume = 1.00 L

So, $Molarity=\frac{Moles}{Volume}=\frac{0.100}{1.00}\ M=0.100\ M$

C = 0.100 M

$2.94=-log(\sqrt{k_a\times 0.100})$

$\log _{10}\left(\sqrt{k_a0.1}\right)=-2.94$

$\sqrt{0.1}\sqrt{k_a}=\frac{1}{10^{2.94}}$

$k_a=1.3\times 10^{-5}$