The `emf` of the following cell is observed to be `0.118V` at `25^(@)C`.
`[Pt,H_(2)(1atm)|HA(100mL 0.1M||H^(o+)(0.1M)|H_(2)(1atm)|Pt]`
a. If `30mL` of `0.2M NaOH` is added to the negative terminal of battery, find the emf of the cell.
b. If `50mL` of `0.2 M NaOH` is added to the negative terminal of battery, find the emf of teh cell.

Explanation

To solve the given problem step by step, let's break it down into parts (a) and (b).

### Part (a): Finding the EMF after adding 30 mL of 0.2 M NaOH

1. **Calculate the initial moles of HA and NaOH:**
- Moles of HA = Volume (L) × Molarity = 0.1 L × 0.1 M = 0.01 moles = 10 millimoles
- Moles of NaOH = Volume (L) × Molarity = 0.03 L × 0.2 M = 0.006 moles = 6 millimoles

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