Question

Determine the pH of each of the following solutions.

Part A:

8.23×10⁻² M HClO4

Express your answer to three decimal places.

Part B:

A solution that is 4.8×10⁻² M in HClO4 and 5.0×10⁻² M in HCl

Express your answer to two decimal places.

Part C:

A solution that is 1.04% HCl by mass (Assume a density of 1.01 g/mL for the solution.)   

Answer 1

PartA

HClO4 is strong acid and completely dissociates into ions

HClO4 <-------> H⁺ + ClO4^-

[HClO4] = 8.23×10^-2M

pH = -log[H⁺]

pH = -log(8.23×10^-2)

pH = 1.08

Part B

HClO4 and HCl both strong acid , so completely dissociates

[HClO4] = 4.8 ×10^-2M

[H⁺] from HClO4 = 4.8 ×10^-2M

[HCl] = 5.0×10^-2M

[H⁺] from HCl= 5.0 ×10^-2M

Total [H⁺ ]= 4.8×10^-2M + 5.0×10^-2​​​​​​M = 9.8×10^-2M

pH = -log(9.8 ×10^-2)

pH = 1.01

Part C

Consider 1L of solution

mass of HCl = 1000ml × 1.01g/ml = 1010g

mass of HCl = (1.04/100)×1010g = 10.504g

Number of moles = mass/molar mass

number of moles of HCl = 10.504g/36.46g/mol = 0.2881mol

molarity = number of moles of solute per liter of solution

molarity of HCl = 0.2881M

molarity of H⁺ = 0.2881M

pH = -log(0.2881)

pH = 0.54