Calculate the pH value of a mixture containing 50 ml of 1 N HCl and 30 ml of 1 N NaOH solution, assuming both to be completely dissociated.

Answer

To calculate the pH value of a mixture containing 50 ml of 1 N HCl and 30 ml of 1 N NaOH, we can follow these steps:

### Step 1: Calculate the number of moles of HCl and NaOH
- **HCl**:
- Normality (N) = 1 N
- Volume (V) = 50 ml = 0.050 L
- Moles of HCl = Normality × Volume = 1 N × 0.050 L = 0.050 moles

- **NaOH**:
- Normality (N) = 1 N
- Volume (V) = 30 ml = 0.030 L
- Moles of NaOH = Normality × Volume = 1 N × 0.030 L = 0.030 moles

### Step 2: Determine the limiting reactant
- The reaction between HCl and NaOH is:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- Since both react in a 1:1 ratio, we can see that:
- Moles of HCl = 0.050 moles
- Moles of NaOH = 0.030 moles

- HCl is in excess because there are more moles of HCl than NaOH.

### Step 3: Calculate the remaining moles of HCl after the reaction
- Moles of HCl remaining = Initial moles of HCl - Moles of NaOH
\[ \text{Remaining moles of HCl} = 0.050 - 0.030 = 0.020 \text{ moles} \]

### Step 4: Calculate the total volume of the solution after mixing
- Total volume = Volume of HCl + Volume of NaOH
\[ \text{Total volume} = 50 \text{ ml} + 30 \text{ ml} = 80 \text{ ml} = 0.080 \text{ L} \]

### Step 5: Calculate the concentration of HCl in the mixture
- Concentration of HCl = Remaining moles of HCl / Total volume
\[ \text{Concentration of HCl} = \frac{0.020 \text{ moles}}{0.080 \text{ L}} = 0.25 \text{ M} \]

### Step 6: Calculate the pH of the solution
- Since HCl is a strong acid, it completely dissociates in solution. Therefore, the concentration of H⁺ ions is equal to the concentration of HCl.
- pH = -log[H⁺]
\[ \text{pH} = -\log(0.25) \approx 0.6021 \]

### Final Answer:
The pH value of the mixture is approximately **0.60**.

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