WEBVTT
00:00:03.320 --> 00:00:10.560
An unknown mass of CaCO3 was dissolved to make up a solution of 80 milliliters.
00:00:11.030 --> 00:00:20.540
Upon titration, it was found that 60 milliliters of four-molar HCl completely neutralized the CaCO3 solution.
00:00:21.410 --> 00:00:25.230
What was the concentration of the CaCO3 solution?
00:00:26.120 --> 00:00:30.030
What mass of CaCO3 was dissolved in the solution.
00:00:30.610 --> 00:00:32.330
Give your answer in grams.
00:00:33.480 --> 00:00:40.550
We can start this problem by writing a balanced chemical equation between HCl and CaCO3.
00:00:41.410 --> 00:00:48.240
When an acid reacts with a carbonate, three products are formed: a salt, water, and carbon dioxide.
00:00:49.480 --> 00:00:58.780
In this example, hydrochloric acid reacts with calcium carbonate to produce calcium chloride, the salt, water, and carbon dioxide.
00:00:59.740 --> 00:01:07.040
Be careful to ensure that the reaction is balanced by adding appropriate coefficients before continuing on with the problem.
00:01:08.250 --> 00:01:14.470
As a titration experiment was performed, we should recall the key equation for solving titration problems.
00:01:15.290 --> 00:01:26.590
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.
00:01:27.610 --> 00:01:32.900
We can make a table to match the values given in the question with the variables of our key equation.
00:01:33.930 --> 00:01:38.920
We will also record the molar ratio of hydrochloric acid to calcium carbonate.
00:01:40.700 --> 00:01:47.300
In the titration experiment, 60 milliliters of a four-molar hydrochloric acid solution were used.
00:01:48.220 --> 00:01:56.040
We can add the concentration to the appropriate box in our table, recognizing that molar and moles per liter are equivalent units.
00:01:57.000 --> 00:02:05.750
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.
00:02:06.680 --> 00:02:10.770
Recognize that 1000 milliliters are equivalent to one liter.
00:02:11.820 --> 00:02:16.970
We can then multiply the 60 milliliters by one liter over 1000 milliliters.
00:02:17.430 --> 00:02:21.540
This gives us a volume of 0.06 liters.
00:02:22.680 --> 00:02:29.960
The hydrochloric acid solution was used to completely neutralize 80 milliliters of a calcium carbonate solution.
00:02:30.740 --> 00:02:40.140
Once again, the volume is given in milliliters and must be converted into liters by multiplying by one liter over 1000 milliliters.
00:02:40.840 --> 00:02:44.210
This gives us the volume of 0.08 liters.
00:02:44.980 --> 00:02:50.270
The first question asked us to determine the concentration of the calcium chloride solution.
00:02:51.690 --> 00:02:56.220
Now that the given values have been filled into the table, we are ready to solve the problem.
00:02:57.090 --> 00:03:08.440
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.
00:03:10.050 --> 00:03:17.910
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.
00:03:18.780 --> 00:03:27.290
Looking at the balanced chemical equation we can see that the molar ratio of hydrochloric acid to calcium carbonate is two to one.
00:03:29.300 --> 00:03:39.050
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.
00:03:39.780 --> 00:03:43.620
We will start the conversion process with moles of hydrochloric acid.
00:03:44.440 --> 00:03:53.640
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.
00:03:55.040 --> 00:04:02.240
We perform the calculation and determine the number of moles of calcium carbonate to be 0.12 moles.
00:04:03.580 --> 00:04:09.800
Next, we can rearrange our key equation in order to solve for the concentration of calcium carbonate.
00:04:10.760 --> 00:04:20.080
We can substitute our calcium carbonate amount and volume and determine the concentration of calcium carbonate to be 1.5 molar.
00:04:21.150 --> 00:04:27.670
The second question asked us to determine the mass of calcium carbonate that was dissolved in the solution.
00:04:28.550 --> 00:04:33.280
In order to calculate the mass, we will need to recognize our second key equation.
00:04:33.920 --> 00:04:48.530
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.
00:04:49.150 --> 00:04:55.710
Be careful not to confuse this capital 𝑀 with the unit molar, also represented by capital M.
00:04:56.680 --> 00:05:06.220
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.
00:05:07.380 --> 00:05:12.660
The molar mass is the sum of the atomic masses of the constituent atoms in the compound.
00:05:13.430 --> 00:05:21.850
For calcium carbonate, that means adding the atomic mass of one calcium atom, one carbon atom, and three oxygen atoms.
00:05:22.260 --> 00:05:25.140
We can find the atomic masses on the periodic table.
00:05:25.700 --> 00:05:30.970
The atomic mass of calcium is 40, carbon 12, and oxygen 16.
00:05:31.760 --> 00:05:36.310
This gives us a molar mass of 100 grams per mole for calcium carbonate.
00:05:37.280 --> 00:05:42.430
Next, we can rearrange our key equation to solve for the mass in grams.
00:05:43.410 --> 00:05:54.660
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.