Answer:
<em>2NaOH+ H2SO4 → Na2SO4 + 2H2O</em>
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HBr reacts with LiOH and forms LiBr and H₂O as the products. The balanced reaction is
LiOH(aq) + HBr(aq) → LiBr(aq) + H₂O(l)
Molarity (M) = moles of solute (mol) / volume of the solution (L)
Molarity of LiOH = 0.205 M
Volume of LiOH = 29.15 mL = 29.15 x 10⁻³ L
Hence,
moles of LiOH = molarity x volume of the solution
= 0.205 M x 29.15 x 10⁻³ L
= 5.97575 x 10⁻³ mol
The stoichiometric ratio between LiOH and HBr is 1 : 1.
Hence,
moles of HBr in 25.0 mL = moles of LiOH added
= 5.97575 x 10⁻³ mol
Hence, molarity of HBr = 5.97575 x 10⁻³ mol / 25.00 x 10⁻³ L
= 0.23903 M
≈ 0.239 M
Hence, the molarity of the HBr is 0.239 M.
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