Question

Determine the pH of each solution.

8.40×10⁻² M HClO4

a solution that is 5.2×10⁻² M in HClO4 and 4.7×10⁻² Min HCl

a solution that is 1.06% HCl by mass (Assume a density of 1.01 g/mL for the solution.)

Answer 1

8.4×10^−2 M HClO4
[H+] = 8.4*10^-2
pH = - log (8.4*10^-2)
pH = 1.08

[H+] of HClO4 = 5.2 x 10^-2 M
[H+] of HCl = 4.7 x 10^-2 M

they are strong acids so [H+]= 5.2 x 10^-2 + 4.7 x 10^-2=9.9 x 10^-2 M

pH = - log 9.9 x 10^-2 =1.004

a solution that is 1.06% HCl by mass (Assume a density of 1.01 g/mL for the solution.)
Take 1.0L of the acid solution:
Mass of solution = 1010g
HCl is 1.06% = 1.06/100*1010 = 10.706 g HCl per litre solution
Molar mass HCl = 36.55g/mol
10.706g = 10.706/36.55 = 0.2929 M HCl
[H+] = 0.2929M
pH = -log 0.2929
pH = 0.53