Calculate the `pH` of a solution which contains `100mL` of `0.1MHCl` and `9.9mL` of `1.0MNaOH`.
Take `(1)/(1099)=9.1xx10xx10^(-4)`& `log9.1=0.96`.

Explanation

To calculate the pH of a solution containing 100 mL of 0.1 M HCl and 9.9 mL of 1.0 M NaOH, we can follow these steps:

### Step 1: Calculate the moles of HCl and NaOH
- **Moles of HCl**:
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{mol} \]

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