QuestionApril 19, 2025

The pH of a 1.5M solution of pentanoic acid (HC_(5)H_(9)O_(2)) is measured to be 2.33. Calculate the acid dissociation constant K_(a) of pentanoic acid . Round your answer to 2 significant digits. K_(a)=square

The pH of a 1.5M solution of pentanoic acid (HC_(5)H_(9)O_(2)) is measured to be 2.33. Calculate the acid dissociation constant K_(a) of pentanoic acid . Round your answer to 2 significant digits. K_(a)=square
The pH of a 1.5M solution of pentanoic acid (HC_(5)H_(9)O_(2)) is measured to be 2.33.
Calculate the acid dissociation constant K_(a) of pentanoic acid . Round your answer to
2 significant digits.
K_(a)=square

Solution
4.4(219 votes)

Answer

K_a = 1.7 \times 10^{-5} Explanation 1. Calculate the concentration of H^+ ions Use the formula for pH: pH = -\log[H^+]. Rearrange to find [H^+]: [H^+] = 10^{-pH}. Substitute pH = 2.33: [H^+] = 10^{-2.33}. 2. Determine the initial and equilibrium concentrations Initial concentration of pentanoic acid is 1.5 M. At equilibrium, [HC_{5}H_{9}O_{2}] = 1.5 - [H^+] and [C_{5}H_{9}O_{2}^-] = [H^+]. 3. Apply the expression for K_a K_a = \frac H^+][C_{5}H_{9}O_{2}^- HC_{5}H_{9}O_{2} . Substitute equilibrium concentrations: K_a = \frac{(10^{-2.33})(10^{-2.33})}{1.5 - 10^{-2.33}}.

Explanation

1. Calculate the concentration of $H^+$ ions<br /> Use the formula for pH: $pH = -\log[H^+]$. Rearrange to find $[H^+]$: $[H^+] = 10^{-pH}$. Substitute $pH = 2.33$: $[H^+] = 10^{-2.33}$.<br />2. Determine the initial and equilibrium concentrations<br /> Initial concentration of pentanoic acid is 1.5 M. At equilibrium, $[HC_{5}H_{9}O_{2}] = 1.5 - [H^+]$ and $[C_{5}H_{9}O_{2}^-] = [H^+]$.<br />3. Apply the expression for $K_a$<br /> $K_a = \frac{[H^+][C_{5}H_{9}O_{2}^-]}{[HC_{5}H_{9}O_{2}]}$. Substitute equilibrium concentrations: $K_a = \frac{(10^{-2.33})(10^{-2.33})}{1.5 - 10^{-2.33}}$.
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