An unknown weak acid has an initial concentration of 0.55 M. What is the pH of the solution if the weak acid also has a pK_a of 5.79?

• A. 0.60
• B. 6.05
• C. 3.02
• D. 5.75

Answer: (C)

Verified step by step guidance

1. Identify the given information: the initial concentration of the weak acid, \([HA]_0 = 0.55\,M\), and the \(pK_a = 5.79\). We need to find the pH of the solution.
2. Recall the relationship between \(pK_a\) and the acid dissociation constant \(K_a\): \(pK_a = -\log K_a\) From this, calculate \(K_a\) using \(K_a = 10^{-pK_a} = 10^{-5.79}\).
3. Set up the expression for the acid dissociation equilibrium: \(HA \rightleftharpoons H^+ + A^-\) Let \(x\) be the concentration of $H^+$ ions produced at equilibrium. Then, \([H^+] = x\), \([A^-] = x\), and \([HA] = 0.55 - x\).
4. Write the expression for the acid dissociation constant \(K_a\) in terms of \(x\): \(K_a = \frac{[H^+][A^-]}{[HA]} = \frac{x^2}{0.55 - x}\) Since \(K_a\) is small (weak acid), assume \(x \ll 0.55\) to simplify the denominator to approximately 0.55.
5. Solve for \(x\) (the \([H^+]\) concentration) using the simplified equation: \(x^2 = K_a \times 0.55\) Then calculate pH using \(pH = -\log x\).