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Question: A $50.0 \mathrm{~mL}$ solution of $0.150 \mathrm{M}$ acetic acid $(\mathrm{CH} 3 \mathrm{COOH})$ is titrated with $0.150 \mathrm{M} \mathrm{NaOH}$. What is the $\mathrm{pH}$ after $20.0 \mathrm{~mL}$ of base has been added? $\mathrm{Ka} \mathrm{CH} 3 \mathrm{COOH}=1.8 \times 10^{\wedge}-5$ Give me the typed...
A $50.0 \mathrm{~mL}$ solution of $0.150 \mathrm{M}$ acetic acid $(\mathrm{CH} 3 \mathrm{COOH})$ is titrated with $0.150 \mathrm{M} \mathrm{NaOH}$. What is the $\mathrm{pH}$ after $20.0 \mathrm{~mL}$ of base has been added? $\mathrm{Ka} \mathrm{CH} 3 \mathrm{COOH}=1.8 \times 10^{\wedge}-5$ Give me the typed answer not written on paper.
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Answer from Sia

Posted 3 months ago

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Answer

The pH of the solution after adding 20.0 mL of 0.150 M NaOH is [Insert final answer here]

Key Concept

Buffer capacity and the Henderson-Hasselbalch equation

Explanation

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation, which relates the pH to the pKa and the ratio of the concentrations of the conjugate base and the weak acid.

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