A buffer solution is prepared by dissolving 1.000 g of sodium acetate (CH;COONA) into 100.00 mL of a 0.100 M solution of acetic acid. Then 2.20 mL of a 10.00 M solution of hydrochloric acid is added to the acetic acid/sodium acetate buffer solution. The K, of acetic acid is 1.8 x 105. What is the pH of the solution after adding HCI? pH =|0.9035 %3D

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### Buffer Solution Preparation and pH Calculation

A buffer solution is prepared by dissolving 1.000 g of sodium acetate (CH₃COONa) into 100.00 mL of a 0.100 M solution of acetic acid. Then 2.20 mL of a 10.00 M solution of hydrochloric acid is added to the acetic acid/sodium acetate buffer solution. The \( K_a \) of acetic acid is \( 1.8 \times 10^{-5} \).

#### Problem Statement
**What is the pH of the solution after adding HCl?**

#### Calculation Result
**pH = 0.9035**

[Image Description]
The image shows a problem related to the preparation of a buffer solution and the calculation of its pH after the addition of hydrochloric acid. A formula or a calculated value \(0.9035\) is provided as the pH of the resulting solution. There are no graphs or diagrams present in the image.

#### Step-by-step Solution:

1. **Calculate the moles of sodium acetate:**
   
   \[
   \text{Molar mass of sodium acetate (CH₃COONa)} = 82.03 \, \text{g/mol}
   \]
   
   \[
   \text{Moles of sodium acetate} = \frac{1.000 \, \text{g}}{82.03 \, \text{g/mol}} = 0.0122 \, \text{mol}
   \]

2. **Moles of acetic acid:**
   
   \[
   \text{Molarity of acetic acid solution} = 0.100 \, \text{M}
   \]
   
   \[
   \text{Volume of acetic acid solution} = 100.00 \, \text{mL} = 0.100 \, \text{L}
   \]
   
   \[
   \text{Moles of acetic acid} = 0.100 \, \text{M} \times 0.100 \, \text{L} = 0.0100 \, \text{mol}
   \]

3. **Calculate the moles of HCl added:**

   \[
   \text{Molarity of HCl solution} = 10.00 \, \
Transcribed Image Text:### Buffer Solution Preparation and pH Calculation A buffer solution is prepared by dissolving 1.000 g of sodium acetate (CH₃COONa) into 100.00 mL of a 0.100 M solution of acetic acid. Then 2.20 mL of a 10.00 M solution of hydrochloric acid is added to the acetic acid/sodium acetate buffer solution. The \( K_a \) of acetic acid is \( 1.8 \times 10^{-5} \). #### Problem Statement **What is the pH of the solution after adding HCl?** #### Calculation Result **pH = 0.9035** [Image Description] The image shows a problem related to the preparation of a buffer solution and the calculation of its pH after the addition of hydrochloric acid. A formula or a calculated value \(0.9035\) is provided as the pH of the resulting solution. There are no graphs or diagrams present in the image. #### Step-by-step Solution: 1. **Calculate the moles of sodium acetate:** \[ \text{Molar mass of sodium acetate (CH₃COONa)} = 82.03 \, \text{g/mol} \] \[ \text{Moles of sodium acetate} = \frac{1.000 \, \text{g}}{82.03 \, \text{g/mol}} = 0.0122 \, \text{mol} \] 2. **Moles of acetic acid:** \[ \text{Molarity of acetic acid solution} = 0.100 \, \text{M} \] \[ \text{Volume of acetic acid solution} = 100.00 \, \text{mL} = 0.100 \, \text{L} \] \[ \text{Moles of acetic acid} = 0.100 \, \text{M} \times 0.100 \, \text{L} = 0.0100 \, \text{mol} \] 3. **Calculate the moles of HCl added:** \[ \text{Molarity of HCl solution} = 10.00 \, \
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