Question

Pure (100%) sulfuric acid has a density of 1.84 g/ml. How many mL of sulfuric acid need to be diluted with water to a total volume of 500 mL to prepare a solution with pH=0.500

Answer 1

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

                                                  = 1840 g

molarity of solution = 1840/98.08 g/mol x 1 L

                               = 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

                                            = 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.