Question

10 mL larutan K-asetat (Mr = 98)

mempunyai pH = 10. Jika Ka CH3COOH = 2 x 10-5, CH3COOK

yang terlarut dalam 200 mL larutannya adalah ... gram. (K = 39; C = 12; H = 1;

O = 16)

Answer 1

PH = 10  
POH  = 4 
[OH] = 10^-4 
[OH] = √Kw/Ka [CH3COOK]
10^-8 = 10^-14/2.10^-5 [CH3COOK] 
[CH3COOK] = 10^-8/5.10^-10 = 20 M 
[CH3COOK] = n/V = n/0,2 = 20 M
n = 4 mol
gr = n.Mr = 4.(2(12)+2(16)+3(1)+39) = 4(98) = 392 gram