The neutralization reaction is:
LiOH + HBr --> LiBr + H2O
Reactants are in 1:1 molar ratio, therefore
"n(LiOH)=n(HBr)=0.200mol\/L\\times0.0400L=0.008mol"
"[LiOH]=\\frac{0.008mol}{0.0126L}=0.635M"
Answer: 0.635 M
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