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.
