Question

Asam lemah HX 0,1 M memiliki pH=2,5. larutan garam KX 0,25 M memiliki pH sebesar...........
tlg dijawab dong,please ^^kalau ribet gak usah pake cara gpp

Answer 1

Dik :
[HX]= 0,1 M
pH HX= 2,5
[H+] = 10^-2,5
10^-2,5= √Ka.M
10^-2,5= √Ka. 10^-1
√10^-5=√Ka. 10^-1
(√10^-5)^2= (√Ka.10^-1)^2
10^-5= Ka.10^-1
Ka=10^-5/10^-1
Ka= 10^-4
garam KX--->Hidrolisis HX dan KOH
Reaksinya
HX+KOH-->KX+H2O

[OH-]= √Kw.Mgaram/Ka
= √10^-14. 25.10^-2/10^-4
=√25.10^-12
=5.10^-6
pOH =-log(5.10^-6)
=6-log5
pH =14-(6-log5)
=8+log5