Consider the neutralization reaction 2 HNO 3 (aq) + Ba (OH) 2 (aq) ⟶ 2 H 2 O (l) + Ba (NO 3) 2 (aq) A 0.120 L sample of an unknown HNO 3 solution required 45.1 mL of 0.200 M Ba (OH) 2 for complete neutralization. What is the concentration of the HNO 3 solution?

The concentration of the HNO3 solution is 0.150 M

Explanation:

Step 1: Data given

Volume of the unknown HNO3 sample = 0.120 L

Volume of the 0.200 M Ba (OH) 2 = 45.1 mL

Step 2: The balanced equation

2HNO3 + Ba (OH) 2 ⟶ Ba (NO3) 2 + 2H2O

Step 3: Calculate moles Ba (OH) 2

moles Ba (OH) 2 = molarity * volume

moles Ba (OH) 2 = 0.200 M * 0.0451 L

moles Ba (OH) 2 = 0.00902 moles

Step 4: Calculate moles of HNO3

For 1 mole of Ba (OH) 2 we need 2 moles of HNO3

For 0.00902 moles of Ba (OH) 2 we need 2*0.00902 = 0.01804 moles

Step 5: Calculate molarity of HNO3

molarity = moles / volume

molarity = 0.01804 / 0.120 L

Molarity = 0.150 M HNO3

The concentration of the HNO3 solution is 0.150 M