Kedalam 1 liter larutan asam asetat 0,2 M, kita masukkan sejumlah NaOH padat hingga pH larutan menjadi 4. jika volume larutan diabaikan serta harga tetapan ionisasi asam asetat adalah 2 x 10^-5, hitunglah massa NaOH yang dimasukkan dalam larutan!
Dzaky111PENYANGGA pH = 4 ---> [H⁺] = 10⁻⁴ n.CH₃COOH = 0,2 mol Ka = 2 x 10⁻⁵ [H⁺] = Ka x (a/g) a/g = [H⁺]/Ka = 10⁻⁴/2x10⁻⁵ = 5/1 CH₃COOH + NaOH ----> CH₃COONa + H₂O M: 0,2 mol x mol R: x mol x mol x mol S: 5x mol - x mol
0,2 - x = 5 x 6x = 0.2 mol x = 0,2/6 = 1/30 mol mol NaOH = x mol = 1/30 mol masa NaOH = mol x Mr = 1/30 x 40 = 4/3 gram
pH = 4 ---> [H⁺] = 10⁻⁴
n.CH₃COOH = 0,2 mol
Ka = 2 x 10⁻⁵
[H⁺] = Ka x (a/g)
a/g = [H⁺]/Ka
= 10⁻⁴/2x10⁻⁵
= 5/1
CH₃COOH + NaOH ----> CH₃COONa + H₂O
M: 0,2 mol x mol
R: x mol x mol x mol
S: 5x mol - x mol
0,2 - x = 5 x
6x = 0.2 mol
x = 0,2/6 = 1/30 mol
mol NaOH = x mol
= 1/30 mol
masa NaOH = mol x Mr
= 1/30 x 40
= 4/3 gram