Question

A) Calculate the pH of a 0.15 M hydrogen fluoride (ka = 4.2 x 10 ^-5)

B) Calculate the pH of a 0.15 M ammonium iodide (ka = 1.8 x 10 ^-5) .

Answer 1

1) Given that

Ka of hydrogen fluoride = 4.2 x 10^-5

concentration C = 0.15 M

We know that

[H+] = (Ka .C)^1/2

pH = -log [H+]

Then,

pH = -log(Ka .C)^1/2

   = - log [ (4.2 x 10^-5₎(0.15 ) ]

= 5.2

Therefore , pH of 0.15 M hydrogen fluoride = 5.2

2)

Given that

Ka of ammonium iodide = 1.8 x 10^-5

concentration C = 0.15 M

We know that

[H+] = (Ka .C)^1/2

pH = -log [H+]

Then,

pH = -log(Ka .C)^1/2

   = - log [ (1.8 x 10^-5₎(0.15 ) ]

= 5.57

Therefore , pH of 0.15 M ammonium iodide = 5.57