Multiple Choice
What is the pH of a solution formed by mixing 140.0 mL of 0.25 M HF with 225.0 mL of 0.31 M NaF, given that the Ka of hydrofluoric acid (HF) is 6.8 x 10^-4?
17. Acid and Base Equilibrium
pH of Weak Acids
Struggling with General Chemistry?
Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
Calculate the pH of a 0.03 M aqueous solution of a weak acid, HA, with a pKa of 3.39.
A
pH = 4.21
B
pH = 2.54
C
pH = 3.39
D
pH = 1.82
Verified step by step guidance1
Identify the given values: the concentration of the weak acid [HA] is 0.03 M, and the pKa of the acid is 3.39.
Use the relationship between pKa and Ka: \( \text{pKa} = -\log_{10}(K_a) \). Rearrange this to find \( K_a \) by calculating \( K_a = 10^{-\text{pKa}} \).
Set up the expression for the dissociation of the weak acid: \( HA \rightleftharpoons H^+ + A^- \). Write the expression for the acid dissociation constant: \( K_a = \frac{[H^+][A^-]}{[HA]} \).
Assume that the initial concentration of \( H^+ \) and \( A^- \) is negligible compared to \( [HA] \), and let \( x \) be the concentration of \( H^+ \) and \( A^- \) at equilibrium. Thus, \( [HA] = 0.03 - x \), \( [H^+] = x \), and \( [A^-] = x \).
Substitute these expressions into the \( K_a \) expression: \( K_a = \frac{x^2}{0.03 - x} \). Since \( x \) is small, approximate \( 0.03 - x \approx 0.03 \). Solve for \( x \) to find \( [H^+] \), and then calculate the pH using \( \text{pH} = -\log_{10}([H^+]) \).
0:44mWatch next
Master ICE Charts of Weak Acids with a bite sized video explanation from Jules
Start learningRelated Videos
Related Practice
pH of Weak Acids practice set
Problem sets built by lead tutorsExpert video explanations