pH = 0.5
pH = -log[H+]
[H+] = 0.316 M
100% sulfuric acid
mass of 1 L of H2SO4 solution = 1000 ml x 1.84 g/ml
                                                Â
= 1840 g
molarity of solution = 1840/98.08 g/mol x 1 L
                             Â
= 18.76 M
To prepare a solution with [H+] = 0.316 M
volume of H2SO4 required = 0.316 M x 500 ml/18.76 M
                                          Â
= 8.42 ml
So we have to take 8.42 ml of 18.76 M (100% H2SO4 solution) and
dilute to 500 ml total volume to get a solution with pH 0.5.
