Question

Harga ph dari 250 ml larutan ca(oh)2 0,005 m adalah...

Answer 1

Ma = n/v
= 0,005/ 0,25
= 0,02 M
b = 2
[OH-] = b×Mb
= 2× 0,02
= 0,04 M
= 4×10-²M
pOH = -log[OH-]
= -log 4×10-²
= 2-log 4
pH = pkw - pOH
= 14-2-log 4
= 12+log 4