Assuming the final solution will be diluted to 1.00 l , how much more hcl should you add to achieve the desired ph?

H = 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