Penyelesaian soal dalam menghitung pH larutan penyangga baik yang berupa penyangga dari asam lemah dan basa konjungasinya maupun penyangga dari basa lemah dan asam konjungasinya
1. Calculate the pH of buffer solution that comprises of 15
mL of 0.1 M CH3COOH solution and 15 mL of CH3COONa if Ka
of CH3COOH = 1.8 x 10-5.
Hitunglah pH larutan bufer yang terdiri atas 15 mL larutan CH3COOH
0,1 M dan 15 mL CH3COONa jika Ka CH3COOH = 1,8 x 10-5.
Answer:
Jawaban:
The number of moles of CH3COOH = 15 mL x 0.1 mmol mL-1.
=
1.5 mmol
Jumlah mol CH3COOH = 15 mL x 0,1 mmol mL-1
=
1.5 mmol
The number of moles of CH3COONa = 15 mL x 0.1 mmol mL-1.
=
1.5 mmol
Jumlah mol CH3COONa = 15 mL x 0,1 mmol mL-1
=
1,5 mmol
Because the number of moles of acid = the number of moles of
conjugate base, thus:
pH = - log Ka
= - log 1.8 x 10-5
= 4.75
Karena jumlah mol asam = jumlah mol basa konjugasinya maka:
pH = - log Ka
= - log 1,8 x 10-5
= 4,75
2. A quantity of 50 mL of 0.1 M (NH4)2SO4
solution added into 50 mL of 0.1 M NH4OH solution.
Sebanyak 50 mL larutan (NH4)2SO4 0,1 M ditambahkan ke dalam 50 mL larutan
NH4OH 0,1 mL.
a. Calculate the pH of the solution after being added with (NH4)2SO4 (Kb of NH4OH = 1.8 x 10-5)
Hitunglah pH
larutan setelah penambahan (NH4)2SO4 Kb NH4OH = 1.8 x 10-5
b. When 5 mL of 0.1 M HCl solution is added into the
mixture, how much is the pH?
Jika ditambahkan 5
mL larutan HCI 0,1 M ke dalam campuran itu, berapakah besar pH yang terjadi?
Answer:
Jawaban:
a. A solution that contains NH4OH and (NH4)2SO4 is a buffer solution because it contains weak base (NH4OH) and its conjugate acid (NH4+) so the pH depends on
the proportion between moles NH4OH and moles of NH4+.
Moles of NH4OH =50 mL x 0.1 mmol mL-1.
= 5 mmol
Larutan yang terdiri
atas NH4OH dan (NH4)2SO4
merupakan larutan bufer karena mengandung basa lemah (NH4OH) dan asam konjugasinya (NH4+) sehingga
besar pH bergantung pada perbandingan antara mol NH4OH dan mol NH4+.
Mol NH4OH = 50mL x 0,1 mmol mL-1 = 5 mmol
In the solution, (NH4)2SO4 will go through the following complete ionization.
Dalam larutannya(NH4)2SO4 akan mengalami
ionisasi sempurna sebagai berikut.
(NH4)2SO4(aq) →2 NH4+aq) + SO42-(aq)
Based on the reaction above, the moles of NH4+, in
the solution is twice of the moles (NH4)2SO4.
moles of NH4+ = 2 x moles of (NH4)2SO4
= 2 × 50 mL x 0.1 mmol mL-1
= 10 mmol
pOH = -log (9 x 10-6) = 6 - log 9
pH = 14 - (6 - log 9)
pH = 8 + log 9 = 8.95
Berdasarkan reaksi di atas, jumlah mol
NH4+ dalam larutan sama dengan dua kali jumlah mol (NH4)2SO4
mol NH4+ = 2 x
mol (NH4)2SO4
= 2 × 50 mL x 0.1
mmol mL-1
= 10 mmol
pOH = -log (9 x 10-6) = 6 - log
9
pH = 14 - (6 - log 9)
pH = 8 + log 9 = 8,95
b. The addition of HCl into the basic buffer solution causes
the reduction of weak base (NH4OH) and increases its conjugate acid
(NH4+).
moles of HCI added = 5 mL x 0.1 mmol mL-1 = 0.5 mmol
Penambahan HCI ke dalam larutan bufer
basa menyebabkan pengurangan basa lemah (NH4OH) dan menambah asam konjugasinya (NH4+).
mol HCI yang ditambahkan = 5 mL x 0,1
mmol mL-1 = 0,5 mmol
Reaction happens according to the equation:
pOH = - log (7.7 x 10-6)
pOH = 6 - log 7.7
pH = 14 – (6 - log 7.7)
pH = 8.89
Reaksi yang terjadi sesuai persamaan:
pOH = -
log (7,7 x 10-6)
pOH = 6 -
log 7.7
pH = 14 –
(6 - log 7,7)
pH = 8,89
Penambahan asam mengakibatkan penurunan pH, tetati relatif kecil dari 8,95 menjadi 8,89.
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