VHCl =0.001L, NHCl =0.1N
VNaCl =0.999L
Vsolution =VHCl +VNaCl =1
∴ Normality of HCl in a resulting solution=0.1×0.001/1 =0.0001N=[H+]
Addition of NaCl does not effect the pH of solution as it is a salt & is neutral.
So, pH=−log[H+] = −log[0.0001] = 4