Hitunglah pH dari 1. 0,1 M H2S 2. 0,016 M CH3COOH Ka =1.10-5 3. 0,6 M NaOH 4. 2,5M NH4OH Kb = 1.10-5 Tolong di jawab ka. Terimakasih

1). H2S termasuk asam lemah, maka menggunakan rumus :
[H+] = √ka x M
= √10-5 x 0,1
= √10-6
=10-³
PH = - log [H+]
= - log 10-³
= 3

2). CH3COOH termasuk asam lemah, maka menggunakan rumus :
[H+] = √ka x M
= √10-5 x 0,016
=√ 16.10-8
=4.10-⁴
PH = - log [H+]
= - log 4.10-⁴
=4 - log 4

3). NaOH termasuk basa kuat, maka menggunakan rumus :
[OH-] = M x valensi basa
= 0,6 x 1
= 0,6 = 6.10-¹
POH = - log [OH-]
= - log 6.10-¹
= 1 - log 6
PH = 14 - POH
= 14 - ( 1 - Log 6 )
= 13 + log 6

4). NH4OH termasuk basa lemah, maka menggunakan rumus :
[OH-] = √kb x M
= √10-5 x 2,5
= √25.10-6
= 5.10-³
POH = - log [OH-]
= - log 5.10-³
= 3 - log 5
PH = 14 - POH
= 14 - (3 - Log 5)
= 11 + log 5

Semoga membantu;))

1. pH = - log [H+]
pH = - log 0,1
pH = 1

2. [H+] =[tex] \sqrt{Ka x Ma} [/tex]
[H+] = [tex] \sqrt{ 10^{-5} x (1,6 x 10^{-1}) } [/tex]
[H+] = 1,264 x [tex] 10^{-3} [/tex]
pH = - log [H+]
pH = - log 1,264 x [tex] 10^{-3} [/tex]
pH = 3 + log1,264
pH = 3 + 0,102
pH = 3,102

3. pOH = - log [OH-]
pOH = - log6 x [tex] 10^{-1} [/tex]
pOH = 1 + log6
pOH = 1,78
pH = 14 - pOH
pH = 14 - 1,78
pH = 12,22

4. [OH-] = ??