Question

Suatu larutan asam lemah [HA] memiliki pH = 4. hitunglah molaritas asam tersebut ! (ka = 10⁻⁵)

Answer 1

PH=-log H+
H+=(ka.M)^1/2 ka=tetapan M=Molaritas
4=-log H+
H+=10^-4
10^-4=(10^-5.M)^1/2
10^-8=10^-5.M
M=10^-3

Answer 2

PH = -log [H+]
[H+] = 4

Ma = [H+]²/Ka
     = 4²/10⁻⁵
      = 16x10⁵