Derek S. answered • 04/25/15
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The first thing you're going to want to do is fill in as many spots as possible for the unknowns. The acid starts in the flask, and you're adding sodium hydroxide (NaOH) into the flask using a burette. So we can start by filling in the "Volume of NaOH added" row by looking at the information given in the other rows.
For Trial 1, the initial reading is 0.00 mL on the burette, and the final reading is 15.00 mL. This means 15.00 mL was dispensed by the burette (since a burette measures how much is dispensed), which should go into the "Volume of NaOH added" row below.
Similarly, you can tell by the differences in volumes how much was added in each subsequent trial. In trial 2, it starts with the 15.00 mL (from the first trial), and ends up with 30.05 mL, so 15.05 mL of NaOH was added in the second trial. In the third trial it increases to 45.10 mL, so another 15.05 mL was added.
The average volume added can next be found by adding the volumes added in each trial and dividing by three.
Unfortunately, this is not the end of the problem, since the concentration has not yet been found. More information is needed to get the final answer, and I don't see it in the information given. To solve a titration problem, pH is usually required in some form, but I can't tell what it is. The C1V1 = C2V2 equation is possibly a hint as to what may be missing.
Since HNO3 is a strong acid and NaOH is a strong base, each mole of NaOH should cancel out one mole of HNO3. With these sort of problems, it usually ends up that one of the trials (likely the last one) ends with a pH of 7, which is when the pH is neutral.
If the pH is neutral at that point, then by figuring out how many total moles of NaOH were added, you can infer how many moles of HNO3 were contained in the initial 10.00 mL of acid placed into the Erlenmeyer flask. After that, it's simply moles divided by volume (in Liters) to give you the concentration in molarity.
For Trial 1, the initial reading is 0.00 mL on the burette, and the final reading is 15.00 mL. This means 15.00 mL was dispensed by the burette (since a burette measures how much is dispensed), which should go into the "Volume of NaOH added" row below.
Similarly, you can tell by the differences in volumes how much was added in each subsequent trial. In trial 2, it starts with the 15.00 mL (from the first trial), and ends up with 30.05 mL, so 15.05 mL of NaOH was added in the second trial. In the third trial it increases to 45.10 mL, so another 15.05 mL was added.
The average volume added can next be found by adding the volumes added in each trial and dividing by three.
Unfortunately, this is not the end of the problem, since the concentration has not yet been found. More information is needed to get the final answer, and I don't see it in the information given. To solve a titration problem, pH is usually required in some form, but I can't tell what it is. The C1V1 = C2V2 equation is possibly a hint as to what may be missing.
Since HNO3 is a strong acid and NaOH is a strong base, each mole of NaOH should cancel out one mole of HNO3. With these sort of problems, it usually ends up that one of the trials (likely the last one) ends with a pH of 7, which is when the pH is neutral.
If the pH is neutral at that point, then by figuring out how many total moles of NaOH were added, you can infer how many moles of HNO3 were contained in the initial 10.00 mL of acid placed into the Erlenmeyer flask. After that, it's simply moles divided by volume (in Liters) to give you the concentration in molarity.
I'm having trouble guessing at what I or the question may be missing, so I would recommend asking your professor for advice after solving as far as possible.
