100 ml of 0.1 M NaCl and 100 ml of 0.2 MNaOH are mixed. What is the change in pH of NaCI solution ?

Answer

To solve the problem step by step, we need to analyze the situation when 100 ml of 0.1 M NaCl is mixed with 100 ml of 0.2 M NaOH.

### Step 1: Determine the initial pH of NaCl solution
- NaCl is a neutral salt formed from a strong acid (HCl) and a strong base (NaOH). Therefore, the pH of a 0.1 M NaCl solution is 7.

### Step 2: Calculate the concentration of NaOH after mixing
- When we mix 100 ml of 0.2 M NaOH with 100 ml of NaCl, the total volume of the solution becomes:
\[ V_{\text{total}} = 100 \, \text{ml} + 100 \, \text{ml} = 200 \, \text{ml} \]
- The number of moles of NaOH in the original solution is:
\[ \text{Moles of NaOH} = 0.2 \, \text{M} \times 0.1 \, \text{L} = 0.02 \, \text{moles} \]
- The concentration of NaOH in the new solution is:
\[ \text{Concentration of NaOH} = \frac{0.02 \, \text{moles}}{0.2 \, \text{L}} = 0.1 \, \text{M} \]

### Step 3: Calculate the concentration of OH⁻ ions
- Since NaOH completely dissociates in solution, the concentration of OH⁻ ions is also 0.1 M.

### Step 4: Calculate the pOH of the solution
- The pOH can be calculated using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] = -\log(0.1) = 1 \]

### Step 5: Calculate the pH of the solution
- The relationship between pH and pOH is given by:
\[ \text{pH} + \text{pOH} = 14 \]
- Therefore, we can find the pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 1 = 13 \]

### Step 6: Determine the change in pH
- The initial pH of the NaCl solution was 7, and after mixing with NaOH, the pH is now 13.
- The change in pH is:
\[ \Delta \text{pH} = \text{Final pH} - \text{Initial pH} = 13 - 7 = 6 \]

### Final Answer
The change in pH of the NaCl solution is **6**.

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