Chapter 15: Problem 85

A solution of acetic acid had the following solute concentrations: \(\left[\mathrm{CH}_{3} \mathrm{COOH}\right]=\) \(0.035 \mathrm{M},\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=7.4 \times 10^{-4} \mathrm{M},\) and \(\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]=7.4 \times 10^{-4} \mathrm{M} .\) Calculate the \(K_{a}\) of acetic acid based on these data.

Short Answer

Expert verified

The \(K_a\) of acetic acid, based on the given data, is \(1.566 \times 10^{-5}\)

Step by step solution

Identify the Required Factor

The problem is requiring the calculation of the acid dissociation constant, \(K_a\), of acetic acid, which is obtained using the concentrations of its various components in the equilibrium reaction.

Define the \(K_a\) Equation

The acid dissociation constant for acetic acid can be expressed and calculated using the equation \(K_a= [H3O+][CH3COO-]/[CH3COOH]\). This equation represents acetic acid's equilibrium. This is where acetic acid dissociates into its positive ion (\(H3O+ \)) and its negative ion (\(CH3COO- \)). The concentration of undissociated acetic acid (\(CH3COOH\)) is in the denominator.

Substitute and Calculate

Now substitute the given concentrations into the equation. The concentration of \(H3O+\) and \(CH3COO-\) are both \(7.4 \times 10^{-4}M\). The concentration of \(CH3COOH\) is \(0.035M\). When you substitute these values in, you get: \( K_a = (7.4 \times 10^{-4}\times 7.4 \times 10^{-4})/0.035\). This simplifies to \(K_a = 1.566 \times 10^{-5}\)

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Short Answer

Step by step solution

Identify the Required Factor

Define the \(K_a\) Equation

Substitute and Calculate

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