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Chapter 15: Problem 37

Propanoic acid, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH},\) is a weak acid. Write the expression defining its acid-ionization constant.

Short Answer

Expert verified
The acid-ionization constant (\(K_a\)) for propanoic acid \(\mathrm{C}_{2}\mathrm{H}_{5} \mathrm{COOH}\) is \[K_a = \frac {[C_2H_5COO^-][H^+]} {[C_2H_5COOH]}\].

Step by step solution

01

Write down the ionization equation for Propanoic acid

In water, propanoic acid ionizes to give propanoate ion and hydronium ion. This ionization can be represented as follows: \[C_2H_5COOH \leftrightarrow C_2H_5COO^- + H^+\]. Here, \(C_2H_5COOH\) is the propanoic acid, \(C_2H_5COO^-\) is the propanoate ion and \(H^+\) is the hydronium ion.
02

Derive the expression for the acid-ionization constant (\(K_a\))

The acid-ionization constant (\(K_a\)) for propanoic acid can be expressed as the ratio of concentrations of the products to the concentration of the reactant. Note that the acid ionization constant does not include concentration of water because water is in large excess. Therefore, the \(K_a\) is as follows: \[K_a = \frac {[C_2H_5COO^-][H^+]} {[C_2H_5COOH]}\]. The square brackets denote the equilibrium concentrations of the ions and acid.

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