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mmol HCl = 50 x 0,1 = 5 mmol
NH4OH + HCl => NH4Cl + H2O
mmol NH4Cl = 5 mmol
mmol NH4OH sisa = 5 mmol
[H+] = 10⁻⁵ x 5/5
[H+] = 10⁻⁵
pH = - log 10⁻⁵ = 5. jawabannya (B)
17. pH = 5
pH = - log [H+]
5 = - log [H+]
[H+] = 10⁻⁵
[H+] = Ka x mmol asam/mmol garam
10⁻⁵ = 10⁻⁵ x 0,1v1/0,4v2
0,1v1 = 0,4v2
v1 = 4v2
v1 = 4/5 x 400 = 320 ml
v2 = 1/5 x 400 = 80 ml
jawabannya (C)
18. pH = 9
pOH = 14 - 9 = 5
[OH-] = 10⁻⁵
mmol NH3 = 100 x 0,2 = 20 mmol
[OH-] = Kb x mmol NH3/(2xmmol (NH4)2SO4 )
10⁻⁵ = 10⁻⁵ x 20/2a
2a = 20
a = 10 mmol
massa (NH4)2SO4 = 10 mmol x 132 mg/mmol = 1320 mg = 1,32 gr
jawabannya (A)
20. larutan garam yang bersifat basa (lakmus merah => biru)
jawabannya (D) HCOONa ( asam lemah dan basa kuat menghasilkan garam besifat basa)
21. yang terhidrolisis total adalah garam dari asam lemah & basa lemah
jawabannya (C) 2 dan 4