How many mL of 0.116 M Ba(OH)₂ are required to completely neutralize 150.0 mL of 0.280 M HNO₃?

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The balanced chemical equation for the reaction between Ba(OH)2 and HNO3 is:$$ \text{Ba(OH)}_2 + 2\text{HNO}_3 \rightarrow \text{Ba(NO}_3)_2 + 2\text{H}_2\text{O} $$From the equation, 1 mole of Ba(OH)2 reacts with 2 moles of HNO3. Given:- Concentration of HNO3 = 0.280 M- Volume of HNO3 = 150.0 mL = 0.150 LMoles of HNO3 = Molarity × Volume = 0.280 M × 0.150 L = 0.042 molesSince 1 mole of Ba(OH)2 reacts with 2 moles of HNO3, the moles of Ba(OH)2 required = 0.042 moles / 2 = 0.021 moles.Given the concentration of Ba(OH)2 is 0.116 M, the volume required can be calculated using the formula:$$ \text{Volume} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.021 \text{ moles}}{0.116 \text{ M}} $$$$ \text{Volume} = 0.181 L $$Since 1 L = 1000 mL, the volume required in mL = 0.181 L × 1000 mL/L = 181 mL.

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