Question

Determine the pH of each solution.

Part A 4.6×10^-2 M HI

Part B 8.43×10⁻² M HClO4

Part C a solution that is 4.3×10⁻² M in HClO4 and 5.6×10⁻² M in HCl

Part D a solution that is 1.11% HCl by mass (Assume a density of 1.01 g/mL for the solution.)

Answer 1

Part A

HI dissociates completely to H+ and I-

so [H+] = 4.6 x 10-2

pH = - log [H+] = - log (4.6 x 10-2) = 1.34

Part B

HClO4 dissociates completely to give H+ and ClO4-

[H+] = 8.43 x 10-2 M

pH = -log [H+] = -log ( 8.43 x 10-2) = 1.07

Part C

Total [H+] = [H+] from HCl + [H+] from HClO4

HCl dissociates completely so [H+] from HCl = 5.6 x 10-2M

HClO4 dissociates completely so [H+] from HClO4 = 4.3 x 10-2M

Total [H+] =  5.6 x 10-2M + 4.3 x 10-2M = 9.9 x 10-2 M

pH = -log [H+] = -log (9.9 x 10-2) = 1

Part D

1.11% HCl means 1.11 g of HCl in 100 g of solution .

Moles of HCl = 1.11/ 36.5 = 0.0304mol = 30.4 mmol

Density of solution = 1.01g/ml

Volume of solution = 100/ 1.01 = 99mL

Molarity of HCl = 30.4/99 =0.307

[H+] = 0.307

pH = -log [H+] = -log (0.307) = 0.51