Koliko je procentna otopina NaOH ako je za neutralizaciju 80 cm3 te otopine utrošeno 100 cm3 otopine sulfatne kiseline koncentracije 1 mol/dm3? Ar Na=23, Ar H=1, Ar O =16
2NaOH+H2SO4------------->Na2SO4+2H2O
n(H2SO4)=c*V=100*10(-3)*1=0.1 mol
n(NaOH):n(H2SO4)=2:1
n(NaOH)=n(H2SO4)*2/1=0.1*2/1=0.2 mol
m(NaOH)=n(NaOH)*Mr=0.2*40=8 g
ro=2.13 g/cm3
mr=ro*V=2.13*80=170.4 g
w(NaOH)=m(NaOH)/mr=8/170.4=0.0469=4.69 %
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Procentnost otopine
2NaOH+H2SO4------------->Na2SO4+2H2O
n(H2SO4)=c*V=100*10(-3)*1=0.1 mol
n(NaOH):n(H2SO4)=2:1
n(NaOH)=n(H2SO4)*2/1=0.1*2/1=0.2 mol
m(NaOH)=n(NaOH)*Mr=0.2*40=8 g
ro=2.13 g/cm3
mr=ro*V=2.13*80=170.4 g
w(NaOH)=m(NaOH)/mr=8/170.4=0.0469=4.69 %