Calculate the pH of a 0.103 M solution of CsHSO4.

Question

Please help. I really do not know where to start.

Answer 1

Start by looking at the compound being dissolved. Here you have CsHSO₄. As this is a salt, it will hydrolyze in water and and so we need to look at the hydrolysis of the HSO₄^- anion.

HSO₄^- + H₂O ==> H₃O⁺ + SO₄^2-

Note that H₃O⁺ is formed so the solution will be slightly acidic (pH > 7)

We need to look up the Ka or pKa for HSO₄^- acting as an acid. I find it to be 1.0x10^-2

We can then write...

Ka = 1.0x10^-2 = [H₃O+][SO₄^2-]/[HSO₄^-]

1.0x10-2 = (x)(x) / 0.103 -x (assume x is small and neglect it for now)

x² = 1.03x10^-3

x = 3.2x10^-4 = [H+] ... assumption that x was small is valid

pH = -log 3.2x10^-4

pH = 3.5