Multiple Choice
Calculate the pH of the solution made by adding 0.50 mol of HOBr and 0.30 mol of KOBr to 1.00 L of water. The value of Ka for HOBr is 2.0 × 10⁻⁹. What is the pH of the solution?
17. Acid and Base Equilibrium
pH of Weak Acids
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Calculate the pH at 25°C of a 0.21 M solution of a weak acid that has Ka = 9.2 x 10⁻⁶.
A
pH = 5.00
B
pH = 4.15
C
pH = 3.12
D
pH = 2.87
Verified step by step guidance1
Identify the given values: the concentration of the weak acid \( [HA] = 0.21 \text{ M} \) and the acid dissociation constant \( K_a = 9.2 \times 10^{-6} \).
Write the expression for the acid dissociation constant \( K_a \) for the weak acid \( HA \): \( K_a = \frac{[H^+][A^-]}{[HA]} \).
Assume that the initial concentration of \( HA \) is 0.21 M and that it dissociates to form \( x \) M of \( H^+ \) and \( A^- \). Therefore, \([H^+] = x\), \([A^-] = x\), and \([HA] = 0.21 - x\).
Substitute these expressions into the \( K_a \) expression: \( 9.2 \times 10^{-6} = \frac{x^2}{0.21 - x} \).
Assume \( x \) is small compared to 0.21, so \( 0.21 - x \approx 0.21 \). Solve the simplified equation \( 9.2 \times 10^{-6} = \frac{x^2}{0.21} \) for \( x \), which represents \([H^+]\), and then calculate the pH using \( \text{pH} = -\log[H^+] \).
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