Question

Berapa gram NaOH (Mr=40) harus dimasukan kedalam 200 ml larutan CH3COOH 0,05 M. Agar diperoleh larutan penyangga dengan pH=4 ka CH3COOH 1x10pangkat min5

Answer 1

PH 4 >> yang sisa asam lemah
[H+] = 10^-4 = (Ka x mol asam lemah )/ mol garam terbentuk
10^-4 = 10^-5 x mol ch3cooh sisa / mol ch3coona
mol ch3cooh sisa/mol ch3coona = 10

mol ch3cooh = 0.05 x 200 = 10 mmol

misal: massa NaOH = X

. NaOH + CH3COOH -> CH3COONa + H2O
M : 1000X/40 mmol 10 mmol
B : 1000X/40 mmol 1000X/40 mmol 1000X/40 mmol
S : - 10-(1000X/40) mmol 1000X/40 mmol

mol ch3cooh sisa/ mol ch3coona = 10
(10-25X) / 25X = 10
10-25X = 250X
10 = 275X
X= 1/275 gram

Massa NaOH = 1/275 gram.