Question

Hitung berapa gram Mg (OH)2 yang terlarut dalam 200 ml larutan Mg(OH)2 yang pH nya = 13! ( Ar Mg = 24, O = 16 , H = 1 )
tolong dibantu ya ka soalnya saya engga tau rumus dan caranya.

Answer 1

PH = 13
pOH = 14 - pH = 1
[OH⁻] = 10⁻¹ = 0,1 M
Mg(OH)₂ --> Mg²⁺ + 2OH⁻
[Mg(OH)₂] = 1/2 x [OH⁻] = 0,05M
M = n/V
n = M V = 0,05 M x 200 mL = 10 mmol = 0,01 mol
massa = n x Mr Mg(OH)₂ = 0,01 x (24 + 2(16+1)) = 0,01 x 58 = 0,58 gram

Answer 2

PH = 14 - pH = 14 - 13 = 1; OH^ - = 10^ -1
OH^ - = b × ( gr ÷ Mr ) × ( 1000 ÷ mL )
10^ -1 = 2 × ( gr ÷ 58 ) × ( 1000 ÷ 200)
gr = 0,58