`10 mL` of `10^(-6) M HCl` solution is mixed with `90mL H_(2)O`. `pH` will change approximately:

Options

A. by 1 unit
B. by 0.3 unit
C. 0.7 unit
D. by 0.1 unit

Answer

To solve the problem step by step, we need to calculate the change in pH when 10 mL of \(10^{-6} M\) HCl is mixed with 90 mL of water.

### Step 1: Calculate the initial concentration of H\(^+\) ions from HCl

Given:
- Volume of HCl solution = 10 mL
- Concentration of HCl = \(10^{-6} M\)

Using the formula for moles:
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 10^{-6} \, M \times 10 \, mL = 10^{-6} \, \text{mol/L} \times 0.01 \, L = 10^{-8} \, \text{mol} \]

Since HCl completely dissociates in water:
\[ \text{Moles of H}^+ = 10^{-8} \, \text{mol} \]

### Step 2: Calculate the total volume after mixing

Total volume after mixing:
\[ \text{Total Volume} = 10 \, \text{mL} + 90 \, \text{mL} = 100 \, \text{mL} = 0.1 \, L \]

### Step 3: Calculate the new concentration of H\(^+\) ions

The concentration of H\(^+\) ions after dilution:
\[ \text{Concentration of H}^+ = \frac{\text{Moles of H}^+}{\text{Total Volume}} = \frac{10^{-8} \, \text{mol}}{0.1 \, L} = 10^{-7} \, M \]

### Step 4: Consider the contribution of H\(^+\) from water

At 25°C, the concentration of H\(^+\) ions from the ionization of water is:
\[ [H^+] = 10^{-7} \, M \]

### Step 5: Calculate the total concentration of H\(^+\) ions

Total concentration of H\(^+\) ions:
\[ \text{Total } [H^+] = 10^{-7} \, M \, (\text{from HCl}) + 10^{-7} \, M \, (\text{from water}) = 2 \times 10^{-7} \, M \]

### Step 6: Calculate the new pH

Using the formula for pH:
\[ \text{pH} = -\log[H^+] \]
\[ \text{pH} = -\log(2 \times 10^{-7}) = -\log(2) - \log(10^{-7}) \approx -0.301 - (-7) \approx 6.699 \]

### Step 7: Calculate the change in pH

Initial pH of the HCl solution:
\[ \text{Initial pH} = 6 \quad (\text{since } 10^{-6} M \text{ HCl gives } pH = 6) \]
New pH after dilution:
\[ \text{New pH} \approx 6.699 \]
Change in pH:
\[ \Delta \text{pH} = \text{New pH} - \text{Initial pH} = 6.699 - 6 = 0.699 \approx 0.7 \]

### Conclusion

The change in pH when 10 mL of \(10^{-6} M\) HCl is mixed with 90 mL of water is approximately **0.7**.

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