. Hitunglah pH larutan berikut.

a. Larutan 500 mL amonia 0,1 M (Kb = 4 × 10–5).

2
#[OH-] = akar Kb x M = 4x10-5 x 10-1 =akar 4x10-6 = 2x10-3
#pOH = -Log[OH-]
= -log(2x10-3) = 3 -log 2
#PH = 14- pOH = 14 - 3-log 2 = 11+log 2
#[OH-] = akar Kb x M = akar 4x10-5 x 10-1 =akar 4x10-6 = 2x10-3
#pOH = -Log[OH-]
= -log(2x10-3) = 3 -log 2
#PH = 14- pOH = 14 - 3-log 2 = 11+log 2

Jawabanmu

2014-01-27T10:59:42+07:00
#[OH-] = akar Kb x M = akar 4x10-5 x 10-1 =akar 4x10-6 = 2x10-3
#pOH = -Log[OH-]
= -log(2x10-3) = 3 -log 2
#PH = 14- pOH = 14 - 3-log 2 = 11+log 2
2014-01-27T11:00:43+07:00
[OH-] = akar Kb x M = akar 4x10^-5 x 10-1 =akar 4x10^-6 = 2x10^-3

pOH = -Log[OH-]
= -log (2x10-3) = 3 -log 2

PH = 14- pOH = 14 - 3-log 2 = 11+log 2