jika 66 mg garam (NH4)2SO4 dilarutkan dalam air hingga 500 mL,kb NH4OH =2×10^-5 dan Mr(NH4)2SO4=132,pH larutan garam tersebut adalah....a.6 b.5 c.4
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Pertanyaan
jika 66 mg garam (NH4)2SO4 dilarutkan dalam air hingga 500 mL,kb NH4OH =2×10^-5 dan Mr(NH4)2SO4=132,pH larutan garam tersebut adalah....a.6 b.5 c.4 d.8 e.9
1 Jawaban
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1. Jawaban Robinn
n = gr / Mr
n = 0,066 / 132
n = 0,0005
M = n / V
M = 0,0005 / 0,5 liter
M = 0,001 M
(NH4)2SO4 → 2NH4+ + SO4^2-
[H+] = √Kw / Kb . n . M
[H+] = √10^-14 / 2 x 10^-5 . 2 . 10^-3
[H+] = √10^-9 x 10^-3
[H+] = √10^-12
[H+] = 10^-6
pH = -log [H+]
pH = -log10^-6
pH = 6