Video Transcript
An unknown mass of CaCO3 was
dissolved to make up a solution of 80 milliliters. Upon titration, it was found that
60 milliliters of four-molar HCl completely neutralized the CaCO3 solution. What was the concentration of the
CaCO3 solution? What mass of CaCO3 was dissolved in
the solution. Give your answer in grams.
We can start this problem by
writing a balanced chemical equation between HCl and CaCO3. When an acid reacts with a
carbonate, three products are formed: a salt, water, and carbon dioxide. In this example, hydrochloric acid
reacts with calcium carbonate to produce calcium chloride, the salt, water, and
carbon dioxide. Be careful to ensure that the
reaction is balanced by adding appropriate coefficients before continuing on with
the problem.
As a titration experiment was
performed, we should recall the key equation for solving titration problems. 𝑛 equals 𝑐𝑣, where 𝑛 represents
the amount in moles, 𝑐 is the concentration in moles per liter, and 𝑣 is the
volume in liters. We can make a table to match the
values given in the question with the variables of our key equation. We will also record the molar ratio
of hydrochloric acid to calcium carbonate.
In the titration experiment, 60
milliliters of a four-molar hydrochloric acid solution were used. We can add the concentration to the
appropriate box in our table, recognizing that molar and moles per liter are
equivalent units. The volume is given in milliliters
but must be converted into liters in order for the liters in the concentration unit
to cancel when solving. Recognize that 1000 milliliters are
equivalent to one liter. We can then multiply the 60
milliliters by one liter over 1000 milliliters. This gives us a volume of 0.06
liters.
The hydrochloric acid solution was
used to completely neutralize 80 milliliters of a calcium carbonate solution. Once again, the volume is given in
milliliters and must be converted into liters by multiplying by one liter over 1000
milliliters. This gives us the volume of 0.08
liters.
The first question asked us to
determine the concentration of the calcium chloride solution. Now that the given values have been
filled into the table, we are ready to solve the problem. We can substitute our hydrochloric
acid concentration and volume into the key equation to determine the number of moles
of hydrochloric acid to be 0.24 moles.
Now that we know the number of
moles of hydrochloric acid used, we can determine the number of moles of calcium
carbonate used in the titration. Looking at the balanced chemical
equation we can see that the molar ratio of hydrochloric acid to calcium carbonate
is two to one. As the molar ratio is not one to
one, we will need to perform a conversion to convert the moles of hydrochloric acid
into moles of calcium carbonate. We will start the conversion
process with moles of hydrochloric acid. We will then multiply by our molar
ratio, written as a fraction with moles of hydrochloric acid in the denominator so
that the units cancel out.
We perform the calculation and
determine the number of moles of calcium carbonate to be 0.12 moles. Next, we can rearrange our key
equation in order to solve for the concentration of calcium carbonate. We can substitute our calcium
carbonate amount and volume and determine the concentration of calcium carbonate to
be 1.5 molar.
The second question asked us to
determine the mass of calcium carbonate that was dissolved in the solution. In order to calculate the mass, we
will need to recognize our second key equation. 𝑛 equals lowercase 𝑚 over capital
𝑀, where 𝑛 is still the amount in moles, lowercase 𝑚 is the mass in grams, and
capital 𝑀 is the molar mass in grams per mole. Be careful not to confuse this
capital 𝑀 with the unit molar, also represented by capital M. We know the amount of moles of
calcium carbonate that were in the original solution, but we need to calculate the
molar mass before we can solve for the mass.
The molar mass is the sum of the
atomic masses of the constituent atoms in the compound. For calcium carbonate, that means
adding the atomic mass of one calcium atom, one carbon atom, and three oxygen
atoms. We can find the atomic masses on
the periodic table. The atomic mass of calcium is 40,
carbon 12, and oxygen 16. This gives us a molar mass of 100
grams per mole for calcium carbonate. Next, we can rearrange our key
equation to solve for the mass in grams. We can substitute the number of
moles of calcium carbonate and the molar mass of calcium carbonate and determine the
mass of calcium carbonate in the solution to be 12 grams.
