Video: Calculating the Concentration of Nitric Acid via Titrating against Known Volume of Barium Hydroxide

A 20 mL sample of aqueous nitric acid (HNO₃) is titrated against a 0.047 M solution of barium hydroxide (Ba(OH)₂). What is the concentration of the nitric acid if an average volume of 55 mL of barium hydroxide solution is needed to reach the end-point of the titration?

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Video Transcript

A 20-milliliter sample of aqueous nitric acid HNO₃ is titrated against a 0.047-molar solution of barium hydroxide Ba(OH)₂. What is the concentration of the nitric acid if an average volume of 55 milliliter of barium hydroxide is needed to reach the end-point of the titration.

This is what the titration might have looked like, with the 20 mL of nitric acid as the titrant in the conical flask and the 0.047-molar barium hydroxide solution as the titrant in the burette. 55 milliliters on average of this barium hydroxide solution is required to neutralize the 20-milliliter sample of nitric acid. The end-point of a titration is the point of neutralization, where all of the aqueous nitric acid has reacted with barium hydroxide.

This point could be monitored by using a pH indicator like phenolphthalein, which would turn a bright pink just after the end point. We’re going to start off by writing the balanced equation for this reaction, calculate the moles of barium hydroxide used, then calculate the number of moles of nitric acid we had to start with. And finally, we’ll calculate the molarity, or concentration, of the original nitric acid solution.

Let’s start by balancing the equation. An acid-base reaction like this will produce a salt plus water. So, nitric acid reacts with barium hydroxide to produce barium nitrate and water. We can add a second nitric acid to balance the nitrate groups in barium nitrate. And we can add a second water molecule to the products to balance the number of hydroxides and H⁺ ions on the reactants side. Now, that we have the balanced reaction equation, we can move on to calculating the number of moles of barium hydroxide in the 55 milliliters of standard solution.

The first thing we need to do is convert the volume of the barium hydroxide solution from milliliters to liters. To do that we can use the relationship that 1000 milliliters is equivalent to a liter. This makes sense because milli equals 10 to the minus three. So, milli is one thousandth of something. Converting 55 milliliters to liters requires us to multiply by one liter per 1000 milliliters, giving us 0.055 liters. We can do the same with our 20 milliliters of nitric acid.

The number of moles of barium hydroxide is equal to its concentration, or molarity, multiplied by the volume. Which is equal to 0.047 molar, or moles per liter, multiplied by 0.055 liters, which gives us 0.002585 moles of barium hydroxide in our 55 milliliters of solution. Now, that we’ve calculated the number of moles of barium hydroxide consumed in the neutralization reaction, we can calculate the number of moles of nitric acid consumed.

For every two equivalent of nitric acid, we require one equivalent of barium hydroxide. So, we can calculate the number of moles of nitric acid by multiplying the number of moles of barium hydroxide by two moles of nitric acid per mole of barium hydroxide. Which is equal to 0.002585 moles of barium hydroxide multiplied by two moles of nitric acid per mole of barium hydroxide, giving us 0.00517 moles of nitric acid. We can put this in our table and observe the number of moles of nitric acid is simply twice that of barium hydroxide.

Now, we can move on to the last step, calculating the molarity of our starting nitric acid solution. The concentration of our solution is equal to the number of moles of nitric acid divided by the volume of solution, giving us 0.00517 moles of nitric acid per 0.020 liters. Giving us a concentration of 0.2585 molar. Since all the values in the question are given to two significant figures, we should give our answer to the same degree of precision. So, our final answer for the starting concentration of the nitric acid in this experiment is 0.26 molar.

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