Question

A student titrated 38.00 mL of a 0.522 M sodium hydroxide solution with 25.00 mL of a 0.785 M hydrochloric acid solution. Which is the limiting reagent? How many mL of excess reactant will remain?

Answer 1

NaOH + HCl --> NaCl + H2O

38.00 mL of a 0.522 M sodium hydroxide
number of mols = molarity * volume of solution
= 38 ml * 0.522 M

= 19.836 milli mols

25.00 mL of a 0.785 M HCl

number of mols = molarity * volume of solution
= 25 ml * 0.785 M

= 19.625 milli mols

NaOH : HCl is in 1:1 mol ratio; that means 1 mol NaOH neutralizes 1 mole HCl

Here number of mols of NaOH present is more than number of mols of HCl present
this makes HCl as limiting reactant . The reaction will stop once all HCl is consumed
*****Limiting reactant = HCl
*****Excess reactant = NaOH

mols of NaOH excess =19.836 milli mols-19.625 milli mols
= 0.211 milli mols
volume of excess NaOH = number of mols/molarity of NaOH
= 0.211 milli mols/0.522 M = 0.404 ml

Volume of excess NaOH = 0.404 ml
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