Do 1 dm3 r-ru NaOH o pH=11 dodano 2.0 cm3 3,5% r-ru HCl o gęstości 1,042 g/ cm3. Oblicz pH otrzymanego roztworu.
NaOH + HCl -----------> NaCl + H20
V=1dm2 V=20 cm3
pH=11 cp=3,5%
pOH=3 d= 1,042g/cm3
cm=0,001 mrr= 20,84 g
nOH-=0,001 ms=0,7294g
ns=0,02
Nadmiar jonów H+ w stosunku do jonów OH- wynosi 0,019
cm=0,0186 mol/dm3
pH= 1,73
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NaOH + HCl -----------> NaCl + H20
V=1dm2 V=20 cm3
pH=11 cp=3,5%
pOH=3 d= 1,042g/cm3
cm=0,001 mrr= 20,84 g
nOH-=0,001 ms=0,7294g
ns=0,02
Nadmiar jonów H+ w stosunku do jonów OH- wynosi 0,019
cm=0,0186 mol/dm3
pH= 1,73