Larutan NaX 0,04 M mempunyai pH = 8 + log 2, Harga Ka HX adalah

Larutan NaX 0,04 M mempunyai pH = 8 + log 2, Harga Ka HX adalah

Jawaban dari pertanyaan kamu adalah 10⁻⁴

pH suatu larutan garam bergantung dari asam dan basa pembentuk garam tersebut. Garam NaX merupakan garam yang terbentuk dari asam lemah dan basa kuat, sehingga sifatnya basa. Besarnya konsentrasi OH⁻ dirumuskan sebagai berikut.
[OH⁻] = √(Kw/Ka.Mg)
dimana
Kw = tetapan air (10⁻¹⁴)
Ka = tetapan asam
Mg = molaritas garam

Dalam soal diketahui.
NaX 0,04 M
pH = 8 + Log 2

Ditanyakan Ka.

Kita tentukan dahulu [OH⁻] dari pH.
pH = 8 + Log 2
pOH = 14 – pH
pOH = 14 – (8 + Log 2)
pOH = 6 – Log 2
[OH⁻] = 2.10⁻⁶

Kemudian kita hitung Ka.
[OH⁻] = √(Kw/Ka.Mg)
2.10⁻⁶ = √(10⁻¹⁴/Ka.0,04)
kedua ruas dikuadratkan.
4.10⁻¹² = 10⁻¹⁴/Ka.0,04
Ka.4.10⁻¹² = 10⁻¹⁴.0,04
Ka = 4.10⁻¹⁶/4.10⁻¹²
Ka = 10⁻⁴

Jadi, jawaban yang tepat adalah 10⁻⁴.

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