To solve the problem step by step, we will calculate the pH after adding 20 ml of 0.09 M NaOH to 30 ml of 0.06 M solution of the protonated form of an anion of acid methanamine (H₂A⁺).
### Step 1: Calculate the number of moles of H₂A⁺
To find the number of moles of H₂A⁺ in the solution, we use the formula:
\[
\text{Number of moles} = \text{Molarity} \times \text{Volume (in L)}
\]
Given:
- Molarity of H₂A⁺ = 0.06 M
- Volume of H₂A⁺ = 30 ml = 0.030 L
Calculating the moles:
\[
\text{Moles of H₂A⁺} = 0.06 \, \text{mol/L} \times 0.030 \, \text{L} = 0.0018 \, \text{mol} = 1.8 \, \text{mmol}
\]
### Step 2: Calculate the number of moles of NaOH
Now, we calculate the number of moles of NaOH added:
Given:
- Molarity of NaOH = 0.09 M
- Volume of NaOH = 20 ml = 0.020 L
Calculating the moles:
\[
\text{Moles of NaOH} = 0.09 \, \text{mol/L} \times 0.020 \, \text{L} = 0.0018 \, \text{mol} = 1.8 \, \text{mmol}
\]
### Step 3: Determine the reaction between H₂A⁺ and NaOH
The reaction between H₂A⁺ and NaOH can be represented as:
\[
\text{H₂A}^+ + \text{OH}^- \rightarrow \text{HA} + \text{H₂O}
\]
Since we have equal moles of H₂A⁺ (1.8 mmol) and NaOH (1.8 mmol), they will completely neutralize each other.
### Step 4: Calculate the remaining species
After the reaction, we will have:
- H₂A⁺: 1.8 mmol - 1.8 mmol = 0 mmol (completely neutralized)
- HA: 1.8 mmol (produced from the reaction)
### Step 5: Calculate the total volume of the solution
The total volume after adding NaOH:
\[
\text{Total Volume} = 30 \, \text{ml} + 20 \, \text{ml} = 50 \, \text{ml} = 0.050 \, \text{L}
\]
### Step 6: Calculate the concentration of HA
The concentration of HA in the solution is:
\[
\text{Concentration of HA} = \frac{\text{Moles of HA}}{\text{Total Volume}} = \frac{0.0018 \, \text{mol}}{0.050 \, \text{L}} = 0.036 \, \text{M}
\]
### Step 7: Use the Henderson-Hasselbalch equation
Since we have a weak acid (HA) and its conjugate base (which is not present here as H₂A⁺ is completely neutralized), we can use the pKa of HA to find the pH.
Given:
- pKa₁ = 2.28 (for H₂A⁺)
- pKa₂ = 9.2 (for HA)
Since we have only HA after the reaction, we will use the average of pKa₁ and pKa₂ to find the pH:
\[
\text{pH} = \frac{pKa_1 + pKa_2}{2} = \frac{2.28 + 9.2}{2} = \frac{11.48}{2} = 5.74
\]
### Final Answer
The pH after the addition of 20 ml of NaOH is **5.74**.
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