The volume of 0.1 M NaOH that would neutralize 2.0 L, 0.050 M HCl would be 1.0 L
Stoichiometric calculation
From the equation of the reaction:
[tex]NaOH + HCl ---- > NaCl + H_2O[/tex]
The mole ratio of NaOH to HCl is 1:1.
Mole of 2L, 0.05 M HCl = 2 x 0.05 = 0.1 mole
Equivalent mole of NaOH = 0.1 mole
Volume of 0.1M, 0.1 mole NaOH = 0.1/0.1 = 1.0 L
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