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

0.456M

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