Soal dan Jawaban pH Larutan Asam dan Basa

Soal dan jawaban cara menentukan hasil perhitungan menentukan derajad keasaman, pH suatu larutan asam dan larutan basa. Soal ini berlaku pada larutan asam kuat dan basa kuat, juga asam dan basa lemah.

1. Tentukan harga pH larutan 0,01 M HCl
   Jawab
   HCl(aq) → H⁺(aq) + Cl^-(aq)
   [H⁺] = [HCl] = 0,001 M = 10^-2 M
   pH = -log [H⁺]
   pH = -log 10^-2
   Jadi pH larutan = 2
2. Berapa pH larutan 0,05 M H₂SO₄ ?
   Jawab
   H₂SO₄(aq) → 2H⁺(aq) + SO₄^2-(aq)
   Dari persamaan reaksi
   [H⁺] = 2 x [H₂SO₄]
   [H⁺] = 2 x 0,005 = 0,1 M = 10^-1 M
   pH = -log [H⁺]
   pH = -log 10^-1
   pH = 1
3. Berapa [H⁺] dalam larutan HNO₃ yang pHnya 2 ?
   Jawab
   HNO₃(aq) → H⁺(aq) + NO^3-(aq)
   pH = -log [H⁺]
   2 = -log [H⁺]
   log [H⁺] = -2
   log [H⁺] = log 10^-2
   [H⁺] = 10^-2 = 0,01 M
4. Tentukan pH larutan 0,01 M NaOH !
   Jawab
   NaOH(aq) → Na⁺(aq) + OH^-(aq)
   [OH^-] = [NaOH]
   [OH^-] = 0,01 M = 10^-2 M
   pOH = -log [OH^-]
   pOH = -log 10^-2
   pOH = 2
   pH = 14 – 2
   pH = 12
5. Berapa [OH^-] yang terdapat dalam larutan KOH yang pHnya 13 ?
   Jawab
   KOH(aq) → K⁺(aq) + OH^-(aq)
   pH =  13
   pOH = 14 – 13 = 1
   pOH = -log [OH^-]
   1 = -log [OH^-]
   log [OH^-] = -1
   log [OH^-] = log 10^-1
   [OH^-] = 10^-1 = 0,1 M
6. Tentukan pH larutan CH₃COOH 0,1 M jika Ka CH₃COOH = 1,8 x 10^-5 !
   Jawab
   Ca = [CH₃COOH] = 0,1 M = 10-1 M
   [H⁺] = √(Ka.Ca
   pH = - log [H⁺]
   pH = - log 1,34.10^-3
   pH = 3 – log 1,34
   pH = 3 – 0,137
   pH = 2,873
7. Berapa pH larutan amonia 0,1 M yang tetapan ionisasinya 1,7.10^-5 ?
   Jawab
   NH₃(g) + H₂O(l) ⇄ NH₄⁺(aq) + OH^-(aq)
   Cb = [NH₃] = 0,1 M = 10^-1 M
   pOH = - log [OH^-]
   pOH = - loh 1,3 . 10^-3
   pOH = 3 – log 1,3
   pOH = 3 – 0,114
   pOH = 2,886
   pH = 14 – pOH
   pH = 14 – 2,886
   pH = 11,114