penentuan [H+] & [OH-]

1) 1 gram CH3COOH (Asam Asetat) dalam 100 mL. Berapakah [H+] ? (Ka = 1,7 x 10-5)
Jawab :
CH3COOH ↔ CH3COO- + H+
[CH3COOH] = (gr ÷ Mr) x (1000 ÷ V)
= (1 ÷ 63) x (1000 ÷ 100)
= 15,87 x 10-2 M, atau (0,16 M)
[CH3COO-] = [H+]
Ka = ([H+] . [CH3COO-] ) ÷ [CH3COOH]
= [H+]2 ÷ [CH3COOH]
[H+] = √ (Ka . [CH3COOH])
= √ (1,7 x 10-5 . 0,16)
= √ (0,27 x 10-5)
= √ (27 x 10-7)
[H+] = 5,2√10-7 atau, 5,2 x 10-7/2 atau, 0,0016

2) 2 gram NH4OH (Amonium Hidroksida) dalam 200 mL. Berapakah [OH-] ? (Kb = 1,7 x 10-5)
Jawab :
NH4OH ↔ NH4+ + OH-
[NH4OH] = (gr ÷ Mr) x (1000 ÷ V)
= (2 ÷ 35) x (1000 ÷ 200)
= 0,29 M
[NH4+] = [OH-]
Ka = ([NH4+] . [OH-]) ÷ [NH4OH]
= [OH-]2 ÷ [NH4OH]
[OH-] = √ (Ka . [NH4OH])
= √ (1,7 x 10-5 . 0,29)
= √ (0,49 x 10-5)
= √ (49 x 10-7)
[OH-] = 7 x 10-7/2 atau, 0,002

3) 3 gram HCN (Asam Sianida) dalam 300 mL. Berapakah [H+] ? (Ka = 1,6 x 10-6)
Jawab :
HCN ↔ H+ + CN-
[HCN] = (gr ÷ Mr) x (1000 ÷ V)
= (3 ÷ 27) x (1000 ÷ 300)
= 0,37 M
[H+] = [CN-]
Ka = ([H+] . [CN-]) ÷ [HCN]
= [H+]2 ÷ [HCN]
[H+] = √ (Ka . [HCN])
= √ (1,6 x 10-6 . 0,37)
= √ (0,59 x 10-6)
= √ (59 x 10-8)
[H+] = 7,68 x 10-4 atau, 0,0008

4) 4 gram HCOOH (Asam Formiat) dalam 400 mL. Berapakah [H+] ? (Ka = 1,6 x 10-6)
Jawab :
HCOOH ↔ H+ + HCOO-
[HCOOH] = (gr ÷ Mr) x (1000 ÷ V)
= (4 ÷ 46) x (1000 ÷ 400)
= 0,22 M
[H+] = [HCOO-]
Ka = ([H+] . [HCOO-]) ÷ [HCOOH]
= [H+]2 ÷ [HCOOH]
[H+] = √ (Ka . [HCOOH])
= √ (1,6 x 10-6 . 0,22)
= √ (0,35 x 10-6)
= √ (35 x 10-8)
[H+] = 5,92 x 10-4 atau, 0,0006