Oblicz pH roztworu zawierającego w 1 dm3 0.020 M Na2SO4 i 0.020 M HCl.
cNa+ = 2 *cNa2SO4 = 2*0.020 M = 0.040 M
cSO42- = 0.020 M
cCl- = 0.020 M
I = ½ (0.040*12 + 0.020*22 + 0.020*12 +0.020*12) =0.080
Dla jonu H+: a*B = 3.0, A = 0.51
ogniwa: K(+)-red | A(-)-utl
elektro: K(-)-red | A(+)-utl
→(+)e.joniz.,powin.e-(-)char.metal.
↓(+)char.metal.(-)e.joniz.,powin.e-,elektrouj.
[H3O+]=10-pH
[HCOOH]
[H3O+]
[HCOO−]
pocz.
0.10
0
zmiana
−4.2 ´ 10-3
+4.2 ´ 10-3
+4.2 ´ 10−3
równow.
0.10 − 4.2 ´ 10−3
= 0.0958 = 0.10
4.2 ´ 10−3
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