tolong dibantu jwb no 6 dan 7

No. 6
[H+] = [(100 x 0,2 x 1)+(100 x 0,1 x 2)]/200
= 0,2
pH = - log 0,2 = 1 - log 2

No. 7
Mol HCl = 100 x 0,2 = 20 mmol
Mol NaOH = 100 x 0,1 = 10 mmol

HCl + NaOH -> NaCl + H2O
m: 20. 10. - -
r: 10. 10. 10. 10
s: 10. - 10. 10
Tersisa asam, shg larutan bersifat asam.
[H+] = jml H+ x Masam = 1 x 10/200
= 5 x 10^-2
pH = - log 5 x 10^-2 = 2 - log 5

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