Larutan ch3coona sebanyak 10 ml memiliki ph=8. Jika ka ch3cooh=1.10-5 garam ch3coona yang terlarut dalam 200 mL larutannya sebanyak (Ar. C=12 g mol H=1 g mol O
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Larutan ch3coona sebanyak 10 ml memiliki ph=8. Jika ka ch3cooh=1.10-5 garam ch3coona yang terlarut dalam 200 mL larutannya sebanyak (Ar. C=12 g mol H=1 g mol O=16 g mol Na=23 g mol
1 Jawaban
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1. Jawaban Devykasih
pH = 8
pOH= 14 - pH = 14 - 8 = 5
[OH-] = 10^-5
CH3COONa -> CH3COO- + Na+
[OH-] = √kw/ka x [CH3COO-]
10^-5 = √10^-14/10^-5 x [CH3COO-]
10^-10 = 10^-9 x [CH3COO-]
[CH3COO-] = 10^-10/10^-9 = 10^-1 M
[CH3COONa] = 1/1 x 10^-1 = 10^-1 M
M = massa/Mr x 1000/V
10^-1 = massa/70 x 1000/200
massa = 1,4 gram