Tentukan gram Ba (CH3COO)2 yg diperlukan membuat 500 ml larutan dgn pH=10 (Ka CH3COOH= 10 -5, Ar Ba =137, C=12, H=1)

PH = 10
pOH = 14 – pH = 14 – 10 = 4
pOH = – log [OH-]
4 = – log [OH-]
[OH-] = 10⁻⁴

[OH-] = √Kw/Ka x M
10⁻⁴ = √10⁻¹⁴/10⁻⁵  x M
10⁻⁸ = 10⁻⁹ x M
M = 10
Ba(CH3COO)2 => Ba²⁺  +  2CH3COO-
       5                    5            10
[Ba(CH3COO)2] = 5 M
Mr Ba(CH3COO)2 = 255
massa Ba(CH3COO)2 = 5 mol/liter x 0,5 liter x 255 gr/mol = 637,5 gram