Jawabanmu

2014-02-04T17:20:36+07:00
1 Asam forminat 0.01 M dan Ka = 1,96 x 10^-4
pH = -log [H+]
pH = -log [ \sqrt{Ka  *Ma}
pH = -log [ \sqrt{0.01 * 1.96 * 10 x^-4} }
pH = -log [0.1 * 1,4 x 10^-2]
pH = -log [14 x 10^-4 ]
pH = -log [4-log 14]

 2. HA 0.1M = HCL 0.012 M
[H+] dari HCl = 1  *[H+]
[H+] = 1* 12* 10^-3
[H+] = 12 * 10^-3
Jika pH sama maka [H+] HA= [H+]HCl 
[H+] =  \sqrt{Ka*Ma}
(12 * 10^-3)^2 =  \sqrt{Ka*(0.1)}^2
144 * 10^-6 = Ka * 0.01
144 * 10^-4 = Ka

1 5 1