JAKA OBJĘTOŚĆ 0,2 MOLOWEGO ROZTWORU NAOH JEST NIEZBĘDNA DO ZOBOJĘTNIENIA 0,5DM3 0,005 MOLOWEGO ROZTWORU H2SO4?
2NaOH + H2SO4 --> Na2SO4 +2H2O
Cm NaOH= 0.2mol/dm3
Cm H2SO4= 0.005mol/dm3
V H2SO4= 0.5dm3
1dm3 H2SO4 - 0.005mol H2SO4
0.5dm3 H2SO4- x
x= 0.0025mol H2S04
2 mole NaOH- 1mol H2SO4
x- 0.0025 mol H2SO4
x= 0.005 mol NaOH
1dm3- 0.2mol NaOH
xdm3 - 0.005mol NaOH
x=0.025dm3 NaOH= 25cm3 NaOH
H2SO4 + 2NaOH => Na2SO4 + 2H2O
n(h2so4)=Cm*V=0.005^0.5=0.0025mol
1mol(h2so4) --- 2 mole(naoh)
0.0025mol --- xmol(naoh)
x=0.005mol(naoh)
Cm=n/V
n=Cm*V
V=n/Cm=0.005/0.2=0.025dm3
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2NaOH + H2SO4 --> Na2SO4 +2H2O
Cm NaOH= 0.2mol/dm3
Cm H2SO4= 0.005mol/dm3
V H2SO4= 0.5dm3
1dm3 H2SO4 - 0.005mol H2SO4
0.5dm3 H2SO4- x
x= 0.0025mol H2S04
2 mole NaOH- 1mol H2SO4
x- 0.0025 mol H2SO4
x= 0.005 mol NaOH
1dm3- 0.2mol NaOH
xdm3 - 0.005mol NaOH
x=0.025dm3 NaOH= 25cm3 NaOH
H2SO4 + 2NaOH => Na2SO4 + 2H2O
n(h2so4)=Cm*V=0.005^0.5=0.0025mol
1mol(h2so4) --- 2 mole(naoh)
0.0025mol --- xmol(naoh)
x=0.005mol(naoh)
Cm=n/V
n=Cm*V
V=n/Cm=0.005/0.2=0.025dm3