Final answer:
The concentration of the barium hydroxide solution is 0.128 M, determined by titrating with a known concentration of nitric acid.
Step-by-step explanation:
The student's question involves the calculation of the concentration of a barium hydroxide solution based on its reaction with a known concentration of nitric acid in a titration experiment. When an aqueous solution of barium hydroxide is titrated with nitric acid, the reaction that takes place is as follows:
Ba(OH)2 (aq) + 2HNO3 (aq) → Ba(NO3)2 (aq) + 2H2O (l)
From the balanced equation, we see that one mole of barium hydroxide reacts with two moles of nitric acid. Using the information that 32.0 mL of 0.200 M nitric acid is required to neutralize 25.0 mL of barium hydroxide solution, we can set up the following stoichiometric calculation:
- The number of moles of HNO3 = volume (L) × concentration (M) = 0.032 L × 0.200 M = 0.0064 moles.
- According to the equation, it takes 2 moles of HNO3 to neutralize 1 mole of Ba(OH)2, so the moles of Ba(OH)2 = 0.0064 moles HNO3 ÷ 2 = 0.0032 moles.
- Now we can find the concentration of Ba(OH)2 by dividing the moles of Ba(OH)2 by its volume in liters: concentration (M) = moles / volume (L) = 0.0032 moles / 0.025 L.
- The concentration of the barium hydroxide solution is therefore 0.128 M.