HNO3(aq)+LiOH(aq)→LiNO3(aq)+H2O(l)
1) n=c×V
n(HNO3)=0.379Lmol×25.00×10−3L=9.475×10−3mol
n(LiOH)=0.573Lmol×12.00×10−3L=6.876×10−3mol
2) As mole ratio n(HNO3):n(LiOH)=1:1 , then n(HNO3) should be equal to n(LiOH)
We have n(HNO3)>n(LiOH),
then LiOH is a limiting reactant.
3) n(HNO3) consumed in the reaction is equal to n(LiOH)
n(HNO3)consumed=6.976×10−3mol
n(HNO3)left=9.475×10−3−6.876×10−3=2.599×10−3mol
4) HNO3(aq)→H+(aq)+NO3−(aq)
as 1 mole of HNO3 gives 1 mole of H+ then n(H+)=n(HNO3)left=2.599×10−3mol
c(H+)=Vn(H+)=25.00×10−3+12×10−32.599×10−3mol=7.024×10−2Lmol
pH=−log10[H+]−log107.024×10−2=1.153
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