31.8 grams of Na2CO3 was dissolved to make a 150-milliliter solution for a titration. The results showed that 44 milliliters of the Na2CO3 solution completely reacted with an unknown concentration of 150 milliliters of H2SO4. What was the concentration of H2SO4?
Let’s start this problem by looking at the first sentence. 31.8 grams of sodium carbonate was dissolved to make a 150-milliliter solution for a titration. This sentence describes the making of a standard solution or a solution with a known concentration. We can use these two pieces of information to find the concentration of the sodium carbonate standard solution. We will also need to recognize two key equations. 𝑛 equals 𝑐𝑣 and 𝑛 equals lowercase 𝑚 over capital 𝑀, where 𝑛 in both equations represents the amount in moles. 𝑐 is the concentration in moles per liter. 𝑣 is the volume in liters. Lowercase 𝑚 is the mass in grams. And capital 𝑀 is the molar mass in grams per mole.
As we need to know the concentration of the standard solution, we can start by setting our two equations equal to one another. Then, we can rearrange the equation to solve for the concentration. We can substitute the mass given in the problem for lowercase 𝑚, but we will need to solve for the molar mass of sodium carbonate. The molar mass can be calculated by adding the atomic masses of the constituent atoms. To find the molar mass of sodium carbonate, we will need two times the atomic mass of sodium plus the atomic mass of carbon plus three times the atomic mass of oxygen. We can find the atomic masses on the periodic table. The atomic mass of sodium is 23, carbon 12, and oxygen 16. We perform the calculation and determine the molar mass of sodium carbonate to be 106 grams per mole. We can substitute this value into our equation.
Lastly, we need to add in the volume. Our volume is currently in milliliters but must be in liters in order to use the equation. Recognize that 1000 milliliters equals one liter. We can multiply the 150 milliliters by one liter over 1000 milliliters. This gives us a volume of 0.15 liters. We can add this value to our equation for 𝑣. We perform the calculation and determine the concentration of the sodium carbonate standard solution to be 2.0 moles per liter. This standard solution of sodium carbonate was used in a titration experiment with sulfuric acid. We should therefore write a balanced chemical equation between the sodium carbonate and the sulfuric acid.
When a carbonate reacts with an acid, three products are formed: a salt, water, and carbon dioxide. When sodium carbonate reacts with sulfuric acid, sodium sulfate, the salt, is formed along with water and carbon dioxide. We will use this balanced chemical equation along with the values given in the problem and our key equation 𝑛 equals 𝑐𝑣 to find the concentration of the sulfuric acid. We can make a table to organize the values given in the question and match them with the variables of our key equation. We will also record the molar ratio of sodium carbonate to sulfuric acid.
Our titration used our standard solution of sodium carbonate, which had a concentration of 2.0 moles per liter, but only used 44 milliliters of the amount that we made. We can add the concentration to the table in the appropriate box. Our volume is given in milliliters and must be converted into liters. As before, we can multiply our volume in milliliters by one liter over 1000 milliliters. This gives us a volume of 0.044 liters. The volume of sulfuric acid used was 150 milliliters. We will convert the 150 milliliters into liters by multiplying by one liter over 1000 milliliters. This gives us a volume of 0.15 liters.
The question is asking us to determine the concentration of sulfuric acid. Now that the given values have been filled in to the table, we are ready to start the next step of the process. We can substitute our sodium carbonate concentration and volume into the key equation to determine the number of moles of sodium carbonate used in the titration to be 0.088 moles. Now that we know the number of moles of sodium carbonate used, we can determine the number of moles of sulfuric acid used in the titration. Looking at our balanced chemical equation, we see that the molar ratio between sodium carbonate and sulfuric acid is one to one. Therefore, if 0.088 moles of sodium carbonate are used in the titration, 0.088 moles of sulfuric acid must also be used.
Next, we can rearrange the key equation to solve for the concentration of sulfuric acid. We can substitute our sulfuric acid amount and volume and determine the concentration of sulfuric acid to be 0.59 molar.