A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution.  *Ka*=1.8x10-5

(a) What is the pH of the buffer ____

(b) Calculate the pH ____ of this solution after 0.020 mol of HCl is added.   

C2H3O2- (aq) + H3O+(aq) <====> HC2H3O2 (aq) + H2O(l)

Keep two decimals for the answer

To solve this buffer problem, we'll use the Henderson-Hasselbalch equation for part (a) and then adjust for the addition of HCl in part (b).

### (a) Calculating the pH of the Buffer

The Henderson-Hasselbalch equation is:

\[
\text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]

Where:
- \([\text{A}^-]\) is the concentration of the conjugate base (sodium acetate, NaC₂H₃O₂).
- \([\text{HA}]\) is the concentration of the weak acid (acetic acid, HC₂H₃O₂).
- \(\text{pKa} = -\log(\text{Ka})\).

Given:
- \( \text{Ka} = 1.8 \times 10^{-5} \)
- \([\text{A}^-] = 0.300 \, \text{mol/L}\)
- \([\text{HA}] = 0.300 \, \text{mol/L}\)

First, we calculate \(\text{pKa}\):

\[
\text{pKa} = -\log(1.8 \times 10^{-5}) \approx 4.74
\]

Now we plug in the values into the Henderson-Hasselbalch equation:

\[
\text{pH} = 4.74 + \log\left(\frac{0.300}{0.300}\right) = 4.74 + \log(1) = 4.74 + 0 = 4.74
\]

So, the pH of the buffer is **4.74**.

### (b) Calculating the pH After Adding HCl

When HCl is added to the buffer, it will react with the acetate ions (\(C_2H_3O_2^-\)) to form acetic acid (\(HC_2H_3O_2\)). The reaction is:

\[
C_2H_3O_2^- + H_3O^+ \rightleftharpoons HC_2H_3O_2 + H_2O
\]

1. **Initial moles before HCl addition**:
   - Moles of \(HC_2H_3O_2\) = 0.300 mol
   - Moles of \(C_2H_3O_2^-\) = 0.300 mol
   - Moles of \(HCl\) added = 0.020 mol

2. **Reaction after adding HCl**:
   - Moles of \(C_2H_3O_2^-\) will decrease by 0.020 mol.
   - Moles of \(HC_2H_3O_2\) will increase by 0.020 mol.

So after the reaction, we have:
- Moles of \(HC_2H_3O_2\) = \(0.300 + 0.020 = 0.320 \, \text{mol}\)
- Moles of \(C_2H_3O_2^-\) = \(0.300 - 0.020 = 0.280 \, \text{mol}\)

Now we recalculate the pH using the Henderson-Hasselbalch equation with the new concentrations:

\[
\text{pH} = 4.74 + \log\left(\frac{0.280}{0.320}\right)
\]

Calculating the log term:

\[
\log\left(\frac{0.280}{0.320}\right) = \log(0.875) \approx -0.056
\]

Now substituting back:

\[
\text{pH} \approx 4.74 - 0.056 \approx 4.68
\]

Thus, the pH of the solution after adding 0.020 mol of HCl is approximately **4.68**.

### Summary of Results
- (a) pH of the buffer: **4.74**
- (b) pH after adding HCl: **4.68**

Do you have any questions or need further details? Here are some related questions for your consideration:
1. How does the pH change with different concentrations of acid and base?
2. What would happen if a strong base was added instead of HCl?
3. How does the buffer capacity relate to the concentrations of acid and conjugate base?
4. What other buffers could be used for a similar pH range?
5. How can you prepare a buffer with a higher pH? 

**Tip:** When working with buffers, it's crucial to remember that their pH can resist change upon the addition of small amounts of acid or base.

A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. Ka=1.8x10-5. (a) What is the pH of the buffer? (b) Calculate the pH of this solution after 0.020 mol of HCl is added.

Explore how to calculate the pH of a buffer solution made from acetic acid and sodium acetate, including the impact of adding HCl. This guide covers the Henderson-Hasselbalch equation and demonstrates the calculations step-by-step.

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