 # Lab 4 Weak Acid Unknown

Satisfactory Essays
304 Words Grammar Plagiarism  Writing  Score Lab 4 Weak Acid Unknown Gabi Mejia
Chem 101
Lab 4: Weak Acid Unknown
Procedure:
When testing the acid, use only between 0.2 g and 0.3 g for each trial (get as precise a measurement as you can). The general procedure is to weigh out your acid, dissolve it in water, add a couple drops of the indicator (phenolphthalein), and then add the sodium hydroxide until you note a color change (from clear to pink). When the color change occurs, you have added enough base to completely react with the acid (the endpoint). You are allowed three trials, and will be graded on accuracy.
Lab Questions:
1.

HA(aq) + NaOH(aq)  NaA(aq) + H2O
2.
Trial 1: .1010 M of NaOH = moles of NaOH / 0.0299 L = 0.0030199 mols of NaOH
Trial 2: .1010 M of NaOH = moles of NaOH / 0.0227 L = 0.0022927 mols of NaOH
Trial 3: .1010 M of NaOH = moles of NaOH / 0.0158 L = 0.0015958 mols of NaOH
After we find the moles of NaOH, we can use that to find the molar mass of each acid.
Trial 1: 0.261g / 0.0030199 = 86.43 mm of acid
Trial 2: 0.233g / 0.0022927 = 101.63 mm of acid
Trial 3: 0.254g / 0.0015958 = 159.17 mm of acid
To find the average molar mass, we just add up the three molar masses we calculated and divide them by three, 86.43 + 101.63 + 159.17 / 3 = 115.74 avg mm
3. Original equation: HA + NaOH  NaA + H2O Diprotic equation: H2A + 2NaOH  Na2A + 2H2O Old mole ratio: 1:1 New mole ratio: 1:2 Example: since the ratio is 1 to 2, the mole ratio is going to decrease by half, the molar mass of the acid is going to increase by half.
4. If someone were to add NaCl to my solid unknown acid the molar mass would be affected by being higher than the molar mass of the actual value we got in lab. We would get the mass of what the acid was as well as the mass of the table salt, therefore, producing a larger number than expected.

## You May Also Find These Documents Helpful

• Good Essays

46.6 mL | 34.8 mL | 46.6-34.8=11.8 mL 46.6-34.8=0.0118 L | 1) Number of moles of NaOH used=0.0118 L of NaOH0.1 M of NaOH…

• 579 Words
• 3 Pages
Good Essays
• Powerful Essays

= [0.0020757 mol / {(23.10 + 25) / 1000} L] * 100 = 4.32 M Trial 4 Mass of KHP transferred = 0.4311 g Volume of Distilled water = 25 mL Volume of NaOH used = 22.60 mL Molar mass of KHP = 204.22 g/mol No. of moles of KHP = Mass of KHP used / Molar mass = 0.4311 g / 204.22 g/mol = 0.0021109 moles Concentration of NaOH = No. of moles / Volume = [0.0021109 mol / {(22.60 + 25) / 1000} L] * 100 = 4.43 M Table: Trail 1 Mass…

• 1631 Words
• 5 Pages
Powerful Essays
• Good Essays

VNaOH = (23.28 mL)(1 L/1000 mL) VNaOH = 0. 02328 L. (c) How many moles of NaOH solution were used in the titration?…

• 635 Words
• 3 Pages
Good Essays
• Good Essays

Using the equation to covert ml to l which is L= mL/1000, we find that 7.1 mL = 0.0071000L and 7ml = 0.0070000L…

• 1045 Words
• 5 Pages
Good Essays
• Satisfactory Essays

Experimental Procedure for melting point: The identification of the melting point of the organic acid was done to find another characteristic of the acid and to test the purity of the recrystallized pure acid. A 2-4 mm layer of unknown sample was placed into a capillary tube sealed on one end. Then the capillary tube was inserted into the side a Bibby Sterlin device. The plateau was set to 200°C on the melting point apparatus. Once the plateau temperature was reached, the sample was watched carefully. When the sample first began to melt and when it was fully melted was recorded. These numbers were the range of the melting point. A slow melting point of the unknown organic acid and a standard sample was completed next. A new plateau was set about 10°C lower than the observed melting point of the unknown sample. This time once the plateau was reached, the heating was no more than 1°C per minute. This gave a much more accurate read of both melting points. If the standard did not melt in the range listed on the label of the bottle, that meant the machine was not working properly. The standard sample and the unknown organic acid melted in their appropriate ranges.…

• 1855 Words
• 18 Pages
Satisfactory Essays
• Powerful Essays

The average titre (19.3mL) was used for the volume of sodium hydroxide, whilst the concentration was 0.1 molL-1. 0.00193 moles of sodium hydroxide were used in this experiment.…

• 1283 Words
• 6 Pages
Powerful Essays
• Better Essays

HA(aq) + H20 (( A- + H3O+ With the following formula the degree to which an acid dissociates (Ka) can be calculated and given a numerical value.…

• 1680 Words
• 7 Pages
Better Essays
• Powerful Essays

2) Measure 10 ml of 0.1 molL-1 of Sodium Hydroxide solution using another measuring cylinder.…

• 1292 Words
• 6 Pages
Powerful Essays
• Satisfactory Essays

How many moles of barium nitrate contain 6.80 x 1024 formula units? 6.80 x 1024 formula units x 1 mol / 6.02 x 1023 formula units = 1.13 x 101…

• 330 Words
• 2 Pages
Satisfactory Essays
• Good Essays

Introduction: The purpose behind (the first step in) this experiment is to show that similarly to week 1, the molarity of an acid or base in solution can be determined (so long as one value’s is known) using titration. In this case though, finding the molarity of the acid used in the reaction is then used to determine the percent of that acid in a vinegar solution and compared to the standard value for % acid present in vinegar. The second part of the experiment was to see if by titrating a solution of NaOH and an unnamed mystery acid, you could find the molar mass of the unknown acid (solving the mystery). It must be understood that the number of moles of the reacting NaOH and the number of moles of the product NaX acid, must both equal (in this case 1:1) in order for the calculation to find the molar mass to work.…

• 976 Words
• 4 Pages
Good Essays
• Good Essays

| 0 | Volume of NaOH (mL) used | 22.6 | 22.2 | 21.7 |…

• 696 Words
• 3 Pages
Good Essays
• Satisfactory Essays

Moles of NaOH = 0.0145 x 0.1 = 0.00145molL-1 Concentration of diluted ethanoic acid = 0.00145…

• 358 Words
• 4 Pages
Satisfactory Essays
• Satisfactory Essays

0.0ml± 0.05ml 5.6ml ± 0.05ml = 5.6 ml ± 0.05ml Data Processing: Determination of the concentration of NaOH(aq)…

• 787 Words
• 4 Pages
Satisfactory Essays
• Satisfactory Essays

9.5 – 5.0 = 4.5. Therefore 4.5 mL of NaOH was needed to neutralize HCl.…

• 271 Words
• 2 Pages
Satisfactory Essays
• Satisfactory Essays

Results and Discussion: Each time completed, PH, and ml of NaOH solution left was recorded in the following table:…

• 435 Words
• 2 Pages
Satisfactory Essays