The ph of a finished water from a water treatment process is 10.74. what amount of .02 n sulfuric acid, in milliliters, is required to neutralize 1 l of finished water assuming alkalinity is zero

The pH of water is given to be 10.74.

This means = pOH = 14 - pH = 3.26

pOH = -log[OH-]= 3.26

[OH-] = 0.00055 M / N

The concentration of H2SO4 is given to be = 0.02 N

volume of water = 1 L

So moles of OH- = number of equivalents of OH-

                     = concentration X volume (L) = 0.00055  X 1 = 0.00055

So number of equivalents of H2SO4 requried = 0.00055

Hence volume of 0.02 N H2SO4 required = Moles / Molarity

                     = 0.00055/0.02 = 0.0275 L = 27.5 mL

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b