Question

use the references to access important values if needed for this question.
in the laboratory a student measures the percent ionization of a 0.413 m solution of acetylsalicylic acid (aspirin), hc₉h₇o₄, to be 2.79%.
calculate value of ( k_{a} ) from this experimental data.
( k_{a}= )

Explanation:

Step1: Calculate the concentration of \(H^+\) ions

The percent ionization formula is \(\text{Percent Ionization}=\frac{[\text{H}^+]}{[\text{HA}]_0}\times100\%\).
Given \([\text{HA}]_0 = 0.413\space M\) and percent ionization \(= 2.79\%\), we can solve for \([\text{H}^+]\):

$$
LATEXBLOCK0
$$

For a weak acid \(HA
ightleftharpoons H^++A^-\), at equilibrium \([\text{H}^+]=[A^-]\) and \([\text{HA}]=[\text{HA}]_0 - [\text{H}^+]\). Since \([\text{H}^+]\) is small compared to \([\text{HA}]_0\) (because percent ionization is small), \([\text{HA}]\approx[\text{HA}]_0\)

Step2: Calculate \(K_a\)

The \(K_a\) expression for \(HA
ightleftharpoons H^++A^-\) is \(K_a=\frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\)
Since \([\text{H}^+]=[A^-]=0.0115\space M\) and \([\text{HA}]\approx0.413\space M\)

$$
LATEXBLOCK1
$$

Answer:

\(3.2\times 10^{-4}\)