Question

Hitunglah [H+] dalam larutan berikut :
1. HCl 0,02 M
2. H2SO4 0,3 M
3. CH3COOH 0,004 M Jika Ka = 1 x 10(pangkat -5)

Answer 1

Rumus asam kuat [H+]= jml H x M
1. [H+]= 1 x 0,02= 0.02
2. [H+]= 2 x 0.3= 0.6
3. asam lemah: 
[H+]= 
=
=2.

Answer 2

[H+] = a. M

1. HCl--> H+  + Cl- a=2
[H+]= 1. 0,02=0,02M

2.H2SO4--> 2H+  + SO4 2-  a=3
[H+]= 2. 0,3=0,6M

3. CH3COOH--> CH3COO- + H+
[H+]= √ka.M = √10^-5 X 4.10^-3= 2.10 pangkat -4