Page 29 - 01 PG Bab 01 Sifat Koligatif.pmd

5. Jawaban: e 9. Jawaban: e
‚
ƒ ‹ ‹
–
mol e = ! = @*
!‡
! @*
= 0,01 mol
Reaksi elektrolisis larutan NaOH e = = 32
Cu
+
–
NaOH(aq) → Na (aq) + OH (aq)

–
–
Katode : 2H O( ) + 2e → 2OH (aq) + H (g) e Ni = = 29,5
2 2
–
–
Anode : 4OH (aq) → 2H O( ) + 4e + O (g)
2 2 ‹ !

=
–
–
mol OH = × mol e = × 0,01 mol = 0,01 mol

×

W = = 1,28 gram

Cu
–3
–
[OH ]= = 0,005 M = 5 × 10 M
*
10. Jawaban: a
–3
pOH = –log 5 × 10
2+
2–
MSO (aq) → M (aq) + SO (aq)
= 3 – log 5 4 4
–
2+
pH = 14 – pOH Katode : M (aq) + 2e → M(s) + – ×2
= 14 – (3 – log 5) Anode : 2H O( ) → O (g) + 4H (aq) + 4e ×1
2
2
––––––––––––––––––––––––––––––––––––––
= 11 + log 5 Reaksi sel : 2M (aq) + 2H O( ) → 2M(s) + O (g) + 4H (aq)
+
2+
2 2
6. Jawaban: a
mol M = $
ANa
r
e= Pada proses penetralan:
valensi
+
mol H = mol OH –
23
= = 23
×
1 +
mol H = mol
i = 15 A
t = 50 menit = 50 × 60 detik = 0,01 mol
××
ei t mol M = 0,005 mol
W=
96.500
"
A M=
×
23 15 50 60 r
"
×
×
= gram
96.500
=
7. Jawaban: d
= 24
W = 0,225 g
11. Jawaban: a
$ ⋅⋅ *
⋅⋅
W= * = Reaksi elektrolisis larutan H SO sebagai berikut.
2 4
2–
+
H SO (aq) → 2H (aq) + SO (aq)
2
4
4
⋅⋅ * + –
0,225 =
→ i · t = 2.412,5 C Katode : 2H (aq) + 2e → H (g)
2

+
–
Anode : 2H O( ) → 4H (aq) + 4e + O (g)
2 2
8. Jawaban: b
Œ
mol O =
2
⋅⋅

* <
‡
W =
−
<
×
mol O =
2
$ ×× <
‡
*
mol × A = –3
r = 6,25 × 10 mol

–
* ⋅ mol e = × mol O
mol = 2
×

–3
= × 6,25 × 10 mol
×
0,1 = *
×
= 2,5 × 10 mol
–2
××
i= mol e – = F
×

– *
mol e =
= 10 ampere
×

2,5 × 10 –2 =

t = 1.930 sekon
28 Elektrolisis