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