Given that
Volume of HCl = 100 ml = 0.1 L
Normality of HCl = 0.1 N
Now we can find
Gram-equivalent of HCl = Normality × Volume
= 0.1 × 0.1 = 0.01
So,
Volume of solution = Volume of H2O + Volume of HCl
= 900 + 100 = 1000 mL
= 100 mL= 1 L
Normality of HCl in resulting solution = 0.01/1 = 0.01 N
pH = −log[H+]
= −log(0.01)
= 2