tentukan ph larutan dari 0.05 mol hf dalam 4L larutan (ka=10pankat-5

HF ===> Asam Lemah

[H^{+}] = \sqrt {Ka} {Ma} 
Ma HF = mol/Volume = 0.05/4 = 0.0125 M

maka:
[H^{+}] = \sqrt {10^{-5}} x {0.0125} = \sqrt {1.25 x 10^{-7}} = \ 5 x 25 x 10^{-9}
[H^{+}] = 5x 10^-4 \sqrt {0.5} 

pH = - log [H^{+}] = - log 5 x 10^-4 \sqrt {0.5} 
pH = 4 - log 5 \sqrt0.5
pH = 4 - log 3.5