Question

Determine the pH of each solution : a) 3.1 X 10 ^-2 M HI b). 0.116 M HClO4 c). a solution that is 5.6×10−2 M in HClO4 and 2.8×10−2 M in HCl d). a solution that is 1.85% HCl by mass (Assume a density of 1.01 g/mL for the solution.)

Answer 1

: a) 3.1 X 10 ^-2 M HI

pH = -log(H) = -log(3.1*10^-2) = 1.50863

b). 0.116 M HClO4

pH = -lG(H+) = -log(0.116) = 0.93554

c). a solution that is 5.6×10−2 M in HClO4 and 2.8×10−2 M in HCl

H+ = 5.6*10^-2 + 2.8*10^-2 = 0.084 M

pH = -log(H) = -log(0.084) = 1.07572

d). a solution that is 1.85% HCl by mass (Assume a density of 1.01 g/mL for the solution.)

assume

m = 100 g in total so

m = 1.85 g ar eHCl

mol = mass/MW = 1.85/36 = 0.05138 mol

mass of water = 100-1.85 = 98.15 g of water

assume D= 1 g /ml then V = 98.15 ml

threfore

M = mol/V = 0.05138/(98.15 /1000) = 0.52348

pH = -log(H) = -log(0.52348) = 0.28109