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? $\mathrm{Ka} \mathrm{CH} 3 \mathrm{COOH}=1.8 imes 10^{\wedge}-5$ 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? $\mathrm{Ka} \mathrm{CH} 3 \mathrm{COOH}=1.8 imes 10^{\wedge}-5$ Give me the typed answer not written on paper.

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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? $\mathrm{Ka} \mathrm{CH} 3 \mathrm{COOH}=1.8 	imes 10^{\wedge}-5$ 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? $\mathrm{Ka} \mathrm{CH} 3 \mathrm{COOH}=1.8 	imes 10^{\wedge}-5$ Give me the typed answer not written on paper.

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Calculate moles of acetic acid and NaOH before reaction**: Before the titration begins, we need to calculate the moles of acetic acid $(\mathrm{CH_3COOH})$ and sodium hydroxide $(\mathrm{NaOH})$ present. The moles of acetic acid are calculated using the volume and molarity of the acetic acid solution, and the moles of NaOH are calculated using the volume and molarity of the NaOH solution added. ‖ ‖ 2 ⋮ **Reaction stoichiometry**: Acetic acid reacts with NaOH in a 1:1 molar ratio according to the equation $\mathrm{CH_3COOH} + \mathrm{NaOH} \rightarrow \mathrm{CH_3COO^-} + \mathrm{H_2O} + \mathrm{Na^+}$. We can use the stoichiometry of the reaction to determine the moles of acetic acid and acetate ion $(\mathrm{CH_3COO^-})$ after the reaction. ‖ ‖ 3 ⋮ **Calculate the concentration of acetic acid and acetate ion**: After the reaction, we calculate the new concentrations of acetic acid and acetate ion in the solution. This is done by subtracting the moles of NaOH added from the moles of acetic acid initially present to find the remaining moles of acetic acid, and assuming that the moles of NaOH added have completely reacted to form an equivalent amount of acetate ion. ‖ ‖ 4 ⋮ **Use the Henderson-Hasselbalch equation**: The pH of the buffer solution can be calculated using the Henderson-Hasselbalch equation: $\mathrm{pH} = \mathrm{pKa} + \log\left(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\right)$, where $\mathrm{pKa} = -\log(\mathrm{Ka})$, $[\mathrm{A^-}]$ is the concentration of the acetate ion, and $[\mathrm{HA}]$ is the concentration of acetic acid. ‖ ‖ 5 ⋮ **Calculate pH**: Using the concentrations of acetic acid and acetate ion, we can calculate the pH of the solution. ‖ ∻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. ⚹◊What is the equation that represents the dissociation of acetic acid (CH3COOH) in water?⍭ Generate me a similar question◊

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