1. What volume in milliliters of 0.100 M HClO₃ is required to neutralize 40.0 mL of 0.140 M KOH? 2. A 25.0 mL solution of HNO₃ is neutralized with 15.7 mL of 0.250 M Ba(OH)₂. What is the concentration of the original HNO₃ solution?

The correct answer is 1) 56 ml and 2) 0.314 M

Explanation:

1. The reaction taking place in the given case is,  

HClO₃ + KOH ⇒ KClO₃ + H2O, the molarity of HClO₃ given is 0.100 M, the molarity of KOH given is 0.140 M and the volume of KOH given is 40 ml, there is a need to find the volume of HClO₃.  

Therefore, the mole of HClO₃ = mole of KOH

= MHClO₃ × VHClO₃ = MKOH × VKOH

= 0.100 M × VHClO₃ = 0.140 M × 40 ml

VHClO₃ = 0.140 M × 40 ml/0.100 M

VHClO₃ = 56 ml.  

2. The reaction taking place is,  

2HNO₃ + Ba(OH)₂ ⇒ Ba(NO₃)₂ + 2H₂O

The volume of HNO₃ given is 25 ml, the molarity of Ba(OH)2 is 0.250 M, the volume of Ba(OH)2 is 15.7 ml, the n or the number of moles of HNO₃ is 2, and the n of Ba(OH)2 is 1, the concentration or M of HNO₃ is,  

M₁V₁/n₁ = M₂V₂/n₂

M₁ × 25/ 2 = 0.25 × 15.7/1

M₁ or molarity of HNO₃ = 0.314 M

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