How to calculate change in ph of water with the addition of strong acid?

My initial pH of water is 6.2, which is a 100 ml of water. I will add 1 ml of .1 M HCl. How do I calculate the new pH?How to calculate change in ph of water with the addition of strong acid?
[H+] = 10^-6.2 = 6.3 x 10^-7 M

moles H+ = 0.100 L x 6.3 x 10^-7 = 6.3 x 10^-8



moles HCl = 0.001 L x 0.1 M = 0.0001



total moles H+ = 6.3 x 10^-8 + 0.0001 = 0.0001

total volume = 101 mL = 0.101 L



[H+] = 0.0001 / 0.101 L= 0.000991 M



pH = 3.00