Answer:
pH = 2.7, therefore amount of H+ needed is 10^-2.7 M
u haf 80 ml of hcl and 90 ml of naoh left, therefore 20 ml of hcl used, and 10ml of naoh used.
mols of H+ = 0.02 x 7 x 10^-2 = 1.4 x 10^-3
mols of OH- = 0.01 x 5 x 10^-2 = 5 x 10^-4
H+ and OH- neutralise each other, so remaining mols of H+ = 9 x 10^-4
u need 10^-2.7 mols of H+, so 10^-2.7 - 9 x 10^-4 = 0.001095 mol
vol of HCL needed = 0.001095 / 7 x 10^-2 = 0.0156 L = 15.6 mL
