Question

16.36 Calculate the pH of an aqueous solution at 25°C that is (a) 0.12 M in HCI, (b) 2.4 M in HNO3, and (c) 3.2 X 10 M in HCIO4 16.54 Calculate the pH of an aqueous solution at 25°C that is 0.34 M in phenol (CH5OH). (K, for phenol - 1.3 x 1010 The pH of an aqueous acid solution is 6.20 at 25°C. Calculate the K, for the acid. The initial acid 16.60 concentration is 0.010 M What is the original molarity of a solution of a weak acid whose K, is 3.5 X 10 and whose pH is 5.26 16.62 at 25°C?

please show steps

Answer 1

Solution:

16.36): All the given acids are strong acids, hence they dissociated completely in aqueous solution.

A) 0.12 M HCl

HCl is a strong acid, hence it dissociates completely.

HCl = H+ + Cl-

Thus,

[H+] = 0.12 M

pH = - log [H+] = - log 0.12 = 0.92

pH = 0.92

B) 2.4 M HNO3

HNO3 is a strong acid.

Therefore,

[H+] = 2.4

pH = - log [H+] = - log 2.4 = 1.32

pH = 1.32

C) 3.2 x 10^-4 M HClO4

HClO4 is a strong acid, hence

[H+] = 3.2 x 10^-4 M

pH = - log [H+]

pH = - 3.2 x 10^-4

pH = 4 - log 3.2 = 4 - 0.51 = 3.49

pH = 3.49