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 litersMolarity = 0.0187mol / (0.010 + 0.0310)L = 0.456M

The answer is !