A 400 mL solution containing 0.0500 M CuSO4 and 0.100 M NaCl is electrolysed between inert electrodes by passing 9650 C of charge (F=96500 C mol^-1).…

The charge passed is $9650/96500=0.100$ mol of electrons. At the cathode, $0.0200$ mol $\mathrm{Cu^{2+}}$ (from $0.400\ \mathrm{L}\times0.0500$) takes $0.0400$ mol of $e^-$; the remaining $0.0600$ mol of electrons reduce water, $2\mathrm{H_2O}+2e^-\to\mathrm{H_2}+2\mathrm{OH^-}$, forming $0.0600$ mol $\mathrm{OH^-}$. At the anode, $0.0400$ mol $\mathrm{Cl^-}$ (from $0.400\times0.100$) is oxidised using $0.0400$ mol of electrons; once chloride is exhausted the remaining $0.0600$ mol of electrons oxidise water, $2\mathrm{H_2O}\to\mathrm{O_2}+4\mathrm{H^+}+4e^-$, producing $0.0600$ mol $\mathrm{H^+}$. The $0.0600$ mol $\mathrm{OH^-}$ and $0.0600$ mol $\mathrm{H^+}$ neutralise exactly, leaving a solution of $\mathrm{Na_2SO_4}$ only. The final $\mathrm{pH}$ is therefore $7$.