Hitung ph dari a) hcl 0,01 M b) ca (OH)2 0 01 M

HCl => H⁺  +  Cl⁻
[H+]=[HCl]=0,01 = 10⁻²
pH = -log[H+]
pH = -log 10⁻² = 2

Ca(OH)₂ => Ca²⁺  +  2 OH⁻
[OH⁻]=2 x [Ca(OH)₂] = 2 x 0,01 = 0,02 = 2 x 10 ⁻²
pOH = -log[OH-]
pOH = -log 2×10⁻²
pOH = 2 – log 2
pH = 14 – pOH
pH = 14 – (2 – log 2)
pH = 12 + log 2 = 12 + 0,3 = 12,3