Question

If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction is complete

Answer 1

• 0.456M

Step-by-step explanation:

1. Calculate the number of moles of each reactant:

• Number of moles = molarity × volume in liter

HNO₃:

• #moles = 0.600M × 10.0mL × 1L/1,000mL = 0.00600 moles

Ba(OH)₂:

• #moles = 0.700M × 31.0mL × 1L/1,000mL = 0.0217 moles

2. Reaction

• 2HNO₃ + Ba(OH)₂ → Ba(NO₃)₂ + 2H₂O

3. Mole ratio

• 2 mol HNO₃ : 1 mol Ba(OH)₂

4. Excess reactant

• As per the mole ratio 0.006 moles of HNO₃ will react with 0.003 moles of Ba(OH)₂.

• Since there are 0.0217 moles of Ba(OH)₂ available it is in excess.

• 0.0217 moles - 0.003 moles = 0.0187 moles will be left after the reaction is complete

5. Molarity of Ba(OH)₂

• Molarity = number of moles / volume of solution in liters
• Molarity = 0.0187mol / (0.010 + 0.0310)L = 0.456M