1.ke dalam larutan 200 mL larutan HCL 0,1 M ditambahkan X ml larutan NH4OH 0,1 M sehingga ph larutan menjadi 9.hitung x jika diketahui Kb NH4OH+1,8x10^-5
2.ke dalam 100 ml larutan ch3cooh 0,5 M ditambahkan 0,8 gram Naoh.berapa ph larutan jika diketahui Ka CH3cooh=1,8x10^-5
pakai jalan ya .-.
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= 8 x 10 ^-1/4 x 10 ^1
= 2 x 10 ^-2
= 0,02 M (molaritas NaOH)
M=gram/Mr x 1000/mL
2 x 10^-2 = 8 x 10^-1/4 x 10^1 . 10^3/mL
2 x 10^-2 = 8 x 10^-2/4 x 10^1 mL
8 x 10^-1 mL = 8 x 10^-2
mL = 8 x 10^-2/8 x 10^-1
= 10^-1
= 0,1 mL(Volume NaOH)
CH₃COONa + NaOH ---> CH₃COONa + H OH
M 100mL.0,5M 0,1mL.0,02M
50 mmol 0,002 mmol - -
B 0,002 mmol 0,002 mmol 0,002 mmol 0,002 mmol
S 49,998 mmol 0 0,002 mmol 0,002 mmol
[H⁺] = Ka . mmol CH₃COOH/mmol CH₃COONa
= 18 x 10^-6.49998 x 10^-3/2 x 10^-3
= 899964 x 10^-9/2 x 10^-3
= 449982 x 10^-6
pH = -log 449982 x 10^-6
= 6-log 449982