Question

Jika 2 L larutan CH3COONa memiliki pH = 8, Ka CH3COOH= 10^-5 dan Mr CH3COONa = 82, jumlah garam yg terlarut adalah...

a) 82 mg c) 164 mg e) 328 mg
b) 41 mg d) 246 mg

Answer 1

Diket dan dit:
*Volume: 2 L
*pH = 8 --> [OH-] = 10^-6
*Mr garam = 82
*Ka CH3COOH = 10^-5

Hanya diket senyawa garamnya, yaitu CH3COONa. Maka, terjadi sistem hidrolisis garam di sini, sehingga:
[OH-] = √Kw x [CH3CONa]/Ka
(10^-6 = √10^-14 x [CH3COONa]/10^-5)^2 ; pangkatkan agar akar hilang
10^-12 = 10^-9 x [CH3COONa]
[CH3COONa] = 10^-3 molar

--> M = n / V (L)
n = 10^-3 molar x 2 L
   = 0.0005 mol
--> n = g/Mr
Massa (g) = 0,002 mol x 82 gram/mol
= 0.041 gram
= 41 mg (B)