jika nilai tatapan ionisasi air kw=4×10pangkat -14 maka ph darilarutan CA (OH)2 0,0002M ADALAH a.3-log 2 b.3-log4 c.11+log 2 d. 11

Ca(OH)2 > Ca^+2 + 2OH^-

0,0002M 0,0004M

Kw = [H^+].[OH^-]

[H^+] = Kw/[OH^-]

[H^+] = 4 x10^-14/0,0004

[H^+] = 10^-10 M

pH = -log(10^-10)

pH = 10 (e)