Please explain step by step Thank you so much! You add 17.00 mL of 0.700 M NaOH to 50.00 mL of pure water, and to this mixture you then add 2.00 mL of 0.100 M HCL. What will be the pH of the resulting solution? mmol NaOH = (17.00m) C0.900m) = 11.90 mmol mmol HCl = ( 2.00m) C0.100m) -0. 200 mmo ! remaining HCI = 11-90-0-200 = 11.70 mmol total volume = 17 mL + 50mL + 2mL = 69.00 mL 11.70 mmol = 0.170 total vol 69.00 mL PH X mmol HCl pho-log(0.190) [PH= 0.77