A titration flask contains 0.00425 moles of an acetic acid sample a. Determine the moles of the base (NaOH) required to completely neutralize the acid. b. If the molarity of the based added from the buret was 0.200 M, determine the volume of the base (in mL) required to reach the end point of this titration. H Type here to search NG Info W

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### Titration Problem and Solution Explanation

#### Problem Statement:

4. A titration flask contains 0.00425 moles of an acetic acid sample.

a. Determine the moles of the base (NaOH) required to completely neutralize the acid.

b. If the molarity of the base added from the buret was 0.200 M, determine the volume of the base (in mL) required to reach the endpoint of this titration.

#### Solution:

**Part a:**

To determine the moles of the base (NaOH) required to completely neutralize the acid, consider that acetic acid (CH₃COOH) reacts with sodium hydroxide (NaOH) in a 1:1 molar ratio. The balanced chemical equation for this reaction is:

\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

From the equation, it is clear that 1 mole of acetic acid reacts with 1 mole of NaOH.

Given:
\[ \text{Moles of acetic acid} = 0.00425 \text{ moles} \]

Since the molar ratio is 1:1:
\[ \text{Moles of NaOH required} = 0.00425 \text{ moles} \]

**Part b:**

To find the volume of the base (NaOH) needed, we use the molarity (M) of the NaOH solution provided:

\[ \text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Liters of solution}} \]

Given the molarity of the NaOH solution:
\[ \text{Molarity (M)} = 0.200 \text{ M} \]

We need to determine the volume (V) of the NaOH solution required to provide 0.00425 moles of NaOH.

Using the formula:
\[ \text{Moles} = \text{Molarity} \times \text{Volume (L)} \]
\[ 0.00425 = 0.200 \times V \]

Solving for V:
\[ V = \frac{0.00425}{0.200} \]
\[ V = 0.
Transcribed Image Text:### Titration Problem and Solution Explanation #### Problem Statement: 4. A titration flask contains 0.00425 moles of an acetic acid sample. a. Determine the moles of the base (NaOH) required to completely neutralize the acid. b. If the molarity of the base added from the buret was 0.200 M, determine the volume of the base (in mL) required to reach the endpoint of this titration. #### Solution: **Part a:** To determine the moles of the base (NaOH) required to completely neutralize the acid, consider that acetic acid (CH₃COOH) reacts with sodium hydroxide (NaOH) in a 1:1 molar ratio. The balanced chemical equation for this reaction is: \[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \] From the equation, it is clear that 1 mole of acetic acid reacts with 1 mole of NaOH. Given: \[ \text{Moles of acetic acid} = 0.00425 \text{ moles} \] Since the molar ratio is 1:1: \[ \text{Moles of NaOH required} = 0.00425 \text{ moles} \] **Part b:** To find the volume of the base (NaOH) needed, we use the molarity (M) of the NaOH solution provided: \[ \text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Liters of solution}} \] Given the molarity of the NaOH solution: \[ \text{Molarity (M)} = 0.200 \text{ M} \] We need to determine the volume (V) of the NaOH solution required to provide 0.00425 moles of NaOH. Using the formula: \[ \text{Moles} = \text{Molarity} \times \text{Volume (L)} \] \[ 0.00425 = 0.200 \times V \] Solving for V: \[ V = \frac{0.00425}{0.200} \] \[ V = 0.
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