please tolong jawab soalnya buat tugas besok

7. H+ = pH - log [H+]
- log [H+] = pH
- log [H+] = 4
- log [H+] = - log 10^-4
[H+] = 10^-4

mol CH3COOH :
= M . V
= 0,2 mol/L . 1 L
= 0,2 mol

mol garam CH3COONa :
= Ka × mol asam / [H+]
= 2.10^-5 × 0,2 mol / 10^-4
= 0,04 mol

Reaksi 1 :
CH3COONa -----> CH3COO- + Na+
awal :0,04 - -
reaksi:0,04 ------0,04-----0,04
akhir :_-
0,04-------0,04

Reaksi 2 :
_______Na+ + OH- ------> NaOH
awal :0,04____________________
reaksi:0,04___0,04_______0,04
akhir :__-_____0,04________0,04

mol NaOH = 0,04 mol


Massa NaOH :
= n . Mr
= 0,04 mol . 40 gram/mol
= 1,6 gram

8. a). (NH4)2SO4:
(NH4)2SO4 ⇔ 2NH4+ + SO4 2-
0,02 0,04 0,02

Yang bisa terhidrolisis hanyalah [NH4+], berarti ambil 0,04 Molar:

[H+] = √Kw x [NH4+]/Kb
= √10^-14 x 4 x 10^-2/10^-5
= 2 x 10^-5,5

pH = 5,5 - log 2

b). OH-= √KW/ka x M
=√10^-14/10^-10 X 10^-2
√10^-6
= 10^-3
OH - = 3 - log1
H + = 14 - 3 - log1
= 11 + log1
pH = 11