tolong bantu saya ya untuk mengerjakan soal ini

4.
[tex]n = \frac{m}{mm} \\ \: \: \: \: \: \: \: \: = \frac{0.4}{40} = 0.01 \: mol[/tex]
M = n/V
= 0,01 / 0,5
= 0,02 M
= 2.10^-2 M

[OH-] = 2. 10^-2
pOH = -log 2.10^-2
= 2-log2
pH = 14-(2-log 2)
= 12+log 2

5. a.
[tex] \alpha = \frac{jumlah \: mol \: terionisasi}{jumlah \: mol \: mula - mula} \\ \: \: \: \: = \frac{0.2}{0.2} = \: 1[/tex]
b. M = n/V
= 0,2/250
= 8 . 10^-4

[H+] = 8. 10^-4
pH = 4-log 8

6.
a. HCl pH = 3
[H+] = 10^-3

b. Mg(OH)2 pH = 9
pOH = 5
[OH-] = 10^-5

c. H3PO4 pH = 2,5
[H+] = 10^-2,5