30 ml of 0.06 M solution of the protonated form of an anion of acid methonime `(H_(2)A^(+))` is treated with 0.09 M NaOH. Calculate pH after addition of 20 ml of base. [`pKa_(1)= 2.28` and `pKa_(2)=9.2`]

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Calculate the millimoles of the protonated form of the anion (H₂A⁺)
Given:
- Volume of H₂A⁺ solution = 30 mL
- Molarity of H₂A⁺ solution = 0.06 M

Millimoles of H₂A⁺ can be calculated using the formula:
\[ \text{Millimoles} = \text{Molarity} \times \text{Volume (L)} \]
Converting volume from mL to L:
\[ 30 \, \text{mL} = 0.030 \, \text{L} \]
Now, calculate the millimoles:
\[ \text{Millimoles of H₂A⁺} = 0.06 \, \text{M} \times 0.030 \, \text{L} = 0.0018 \, \text{mol} = 1.8 \, \text{mmol} \]

### Step 2: Calculate the millimoles of NaOH added
Given:
- Volume of NaOH = 20 mL
- Molarity of NaOH = 0.09 M

Using the same formula:
\[ \text{Millimoles of NaOH} = 0.09 \, \text{M} \times 0.020 \, \text{L} = 0.0018 \, \text{mol} = 1.8 \, \text{mmol} \]

### Step 3: Determine the reaction between H₂A⁺ and NaOH
The reaction can be represented as:
\[ H₂A^+ + OH^- \rightarrow HA + H_2O \]
Since we have equal millimoles of H₂A⁺ and NaOH (1.8 mmol each), they will completely react with each other.

### Step 4: Calculate the remaining species after the reaction
After the reaction:
- Millimoles of H₂A⁺ left = 1.8 mmol - 1.8 mmol = 0 mmol
- Millimoles of HA formed = 1.8 mmol

### Step 5: Use the Henderson-Hasselbalch equation to find pH
Since we now have only HA (the deprotonated form) and no H₂A⁺ left, we can use the average of the two pKa values to find the pH:
\[ \text{pH} = \frac{pK_a1 + pK_a2}{2} \]
Substituting the given pKa values:
\[ \text{pH} = \frac{2.28 + 9.2}{2} = \frac{11.48}{2} = 5.74 \]

### Final Answer
The pH after the addition of 20 mL of NaOH is **5.74**.

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