In a titration experiment, a 12.5 ml sample of 0.0175M Ba(OH)2 just neutralized 14.5 ml of HNO3 solution. Calculate the molarity of the HNO3 solution.
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In a titration experiment, a 12.5 ml sample of 0.0175M Ba(OH)2 just neutralized 14.5 ml of HNO3 solution. Calculate the molarity of the HNO3 solution.
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Answer:
Explanation:
To calculate the molarity of the HNO3 solution, you can use the concept of stoichiometry and the balanced chemical equation between Ba(OH)2 and HNO3. The balanced equation is:
2 Ba(OH)2 + 2 HNO3 → Ba(NO3)2 + 2 H2O
From the equation, you can see that it takes 2 moles of HNO3 to neutralize 2 moles of Ba(OH)2.
Given:
Volume of Ba(OH)2 solution = 12.5 ml
Molarity of Ba(OH)2 solution = 0.0175 M
Volume of HNO3 solution = 14.5 ml
First, let's calculate the number of moles of Ba(OH)2 that reacted:
Moles of Ba(OH)2 = Molarity × Volume
Moles of Ba(OH)2 = 0.0175 M × 12.5 ml = 0.21875 moles
Since the stoichiometric ratio between Ba(OH)2 and HNO3 is 1:2, the moles of HNO3 that reacted will be twice the moles of Ba(OH)2.
Moles of HNO3 = 2 × Moles of Ba(OH)2 = 2 × 0.21875 moles = 0.4375 moles
Finally, calculate the molarity of HNO3:
Molarity of HNO3 = Moles of HNO3 / Volume of HNO3 solution (in liters)
Molarity of HNO3 = 0.4375 moles / (14.5 ml / 1000) L = 30.17 M
Therefore, the molarity of the HNO3 solution is approximately 30.17 M.