We can answer this by looking directly at the chemical reaction that occurs during charging. During discharge, the reaction at both plates is: \[ \text{Pb} + \text{PbO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow 2\text{PbSO}_4 + 2\text{H}_2\text{O}. \] Charging drives this reaction in reverse: \[ 2\text{PbSO}_4 + 2\text{H}_2\text{O} \rightarrow \text{Pb} + \text{PbO}_2 + 2\text{H}_2\text{SO}_4. \] This reverse reaction consumes water and regenerates sulfuric acid, directly increasing the acid concentration in the electrolyte. Let's check each option against this reaction.
The reversed charging reaction confirms that the electrolyte becomes stronger.
Therefore, the correct answer is stronger.