Answer to Question #147531 in General Chemistry for gia

Question

What is the pH of 250.0 mL of an aqueous solution containing 0.615 g of the strong acid trifluoromethane sulfonic acid (CF3SO3H)?

Answer 1

CF₃SO₃H + H₂O = [CF₃SO₃]^- + H₃O⁺  complete dissociation as a strong acid

n=m/M=0.615g / 150g/mol = 0.0041 mol

c(CF₃SO₃H)=[H⁺]=n/V=0.0041/0.25=0.0164 mol/L

pH=-lg[H⁺]=1.8

pH=1.8