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maka [H⁺] = 10⁻⁵
LARUTAN PENYANGGA
* Mencari mol CH₃COOH
mol = M . Volume (dalam liter)
mol = 0,1 x 0,5
mol = 0,05 mol
* mencari mol CH₃COONa
[H⁺] = mol asam/mol garam x Ka
10⁻⁵ = mol CH₃COOH/mol CH₃COONa . 10⁻⁵
0,05/mol CH₃COONa = 10⁻⁵/10⁻⁵
0,05/mol CH₃COONa = 1
mol CH₃COONa = 0,05 mol
* mencari massa CH₃COONa
massa = mol x Mr
massa = 0,05 x 82
massa = 4,1 gram
Jadi, massa gram CH₃COONa yang harus ditambahkan adalah 4,1 gram