10^-3 V + 10^-5 V = N 2V. N= 10^-3/2 + 10^-5/2N= 100×10^-5/2 + 10^-5/2N= 101/2 × 10^-5N= 50×10^-5.pH = --log ( 50×10^-5)pH= 5-- log 50pH= 3.3

Understanding pH Calculation
To calculate the pH of a solution obtained by mixing equal volumes of two solutions with different pH values, we need to use the concept of hydrogen ion concentration.

Step 1: Calculate Hydrogen Ion Concentrations
- The formula for hydrogen ion concentration \([H^+]\) is given by:
\[
[H^+] = 10^{-\text{pH}}
\]
- For the solution with pH = 3:
\[
[H^+] = 10^{-3} = 0.001 \, \text{M}
\]
- For the solution with pH = 5:
\[
[H^+] = 10^{-5} = 0.00001 \, \text{M}
\]

Step 2: Calculate Total Hydrogen Ion Concentration
- When equal volumes are mixed, the total concentration of hydrogen ions is the average of the two concentrations:
\[
[H^+]_{\text{total}} = \frac{[H^+]_1 + [H^+]_2}{2}
\]
- Thus, we calculate:
\[
[H^+]_{\text{total}} = \frac{0.001 + 0.00001}{2} = \frac{0.00101}{2} = 0.000505 \, \text{M}
\]

Step 3: Calculate the New pH
- Finally, we calculate the new pH using the total hydrogen ion concentration:
\[
\text{pH} = -\log([H^+]_{\text{total}})
\]
- Substituting the values:
\[
\text{pH} = -\log(0.000505) \approx 3.3
\]

Conclusion
By mixing equal volumes of solutions with pH 3 and pH 5, the resulting solution has a pH of approximately 3.3. This demonstrates how pH is not linear and depends on the logarithmic scale of hydrogen ion concentrations.