Multiple Choice
A 30.00-mL sample of 0.125 M HCOOH is being titrated with 0.175 M NaOH. What is the pH after 15.0 mL of NaOH has been added? Ka of HCOOH = 1.8 × 10⁻⁴
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18. Aqueous Equilibrium
Titrations: Weak Acid-Strong Base
Multiple Choice
What is the pH of a 35.0-mL sample of 0.150 M acetic acid (CH3COOH, Ka=1.8×10⁻⁵) before any 0.150 M NaOH solution is added?
A
4.75
B
1.00
C
7.00
D
2.87
Verified step by step guidance1
Start by writing the dissociation equation for acetic acid: \( \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \).
Use the expression for the acid dissociation constant \( K_a \) for acetic acid: \( K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \).
Set up an ICE table (Initial, Change, Equilibrium) to determine the concentrations of the species at equilibrium. Initially, \([\text{CH}_3\text{COOH}] = 0.150 \text{ M}\), \([\text{CH}_3\text{COO}^-] = 0\), and \([\text{H}^+] = 0\).
Assume that \( x \) is the change in concentration for \([\text{CH}_3\text{COO}^-]\) and \([\text{H}^+]\), so at equilibrium \([\text{CH}_3\text{COOH}] = 0.150 - x\), \([\text{CH}_3\text{COO}^-] = x\), and \([\text{H}^+] = x\).
Substitute these equilibrium concentrations into the \( K_a \) expression: \( 1.8 \times 10^{-5} = \frac{x^2}{0.150 - x} \). Solve for \( x \), which represents \([\text{H}^+]\), and then calculate the pH using \( \text{pH} = -\log[\text{H}^+] \).
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