Question

use the references to access important values if needed for this question. the hydronium ion concentration of an aqueous solution of 0.413 m acetylsalicylic acid (aspirin), hc₉h₇o₄, (kₐ = 3.00×10⁻⁴) is h₃o⁺ = m. an error has been detected in your answer. check for typos, miscalculations etc. before submitting your answer. submit answer retry entire group 6 more group attempts remaining

Explanation:

Step1: Write the dissociation equation

$$\ce{HC9H7O4 + H2O <=> H3O+ + C9H7O4-}$$
Let \(x = [\ce{H3O+}]=[\ce{C9H7O4-}]\) and \([\ce{HC9H7O4}]=0.413 - x\). Since \(K_a\) is small (\(K_a=3.00\times 10^{-4}\)), we can assume \(0.413 - x\approx0.413\)

Step2: Write the \(K_a\) expression

$$K_a=\frac{[\ce{H3O+}][\ce{C9H7O4-}]}{[\ce{HC9H7O4}]}$$
Substitute \(K_a = 3.00\times 10^{-4}\), \([\ce{H3O+}]=x\), \([\ce{C9H7O4-}]=x\) and \([\ce{HC9H7O4}]=0.413\) into the \(K_a\) expression:
$$3.00\times 10^{-4}=\frac{x\times x}{0.413}$$
$$x^{2}=3.00\times 10^{-4}\times0.413$$
$$x^{2}=1.239\times 10^{-4}$$
$$x=\sqrt{1.239\times 10^{-4}}$$
$$x = 0.0111$$

Answer:

\(0.0111\)