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A local company sent you their green alternative for window cleaner to be tested for percent (w/v) acetic acid content. For your experiment, you first standardized your NaOH titrant with 0.8053 g of (99.80 % purity) KHP. You used 40.60 mL of NaOH for your standardization. After that you then analyzed a 10.00 mL sample and found that you needed 43.20 mL NaOH to reach the end point.
Summary of results:
Standardization |
Sample analysis |
||
KHP Weight (g) |
0.8053 g |
Volume of sample |
50.00 mL |
Purity |
99.80% |
NaOH (mL) used |
33.20 mL |
NaOH (mL) used |
40.60 mL |
|
|
Determine the following:
- Molarity of NaOH
- % (w/v) acetic acid
Step by step
Solved in 2 steps with 2 images
- A local company sent you their green alternative for window cleaner to be tested for percent (w/v) acetic acid content. For your experiment, you first standardized your NaOH titrant with 0.8053 g of (99.80 % purity) KHP. You used 40.60 mL of NaOH for your standardization. After that you then analyzed a 50.00 mL sample and found that you needed 33.20 ml NaOH to reach the end point. Summary of results: Standardization Sample analysis KHP Weight (g) Purity NaOH (mL) used Determine the following: 0.8053 g Volume of sample 50.00 mL 99.80% NaOH (mL) used 33.20 mL 40.60 mL a. Molarity of NaOH b. % (w/v) acetic acidA local company sent you their green alternative for window cleaner to be tested for percent (w/v) acetic acid content. For your experiment, you first standardized your NaOH titrant with 0.8053 g of (99.80 % purity) KHP. You used 40.60 mL of NaOH for your standardization. After that you then analyzed a 50.00 mL sample and found that you needed 33.20 mL NaOH to reach the end point. Find the % (w/v) acetic acid.A local company sent you their green alternative for window cleaner to be tested for percent (w/v) acetic acid content. For your experiment, you first standardized your NaOH titrant with 0.8053 g of (99.80 % purity) KHP. You used 40.60 mL of NaOH for your standardization. After that you then analyzed a 10.00 mL sample and found that you needed 43.20 mL NaOH to reach the end point. Determine the following: Molarity of NaOH % (w/v) acetic acid
- A 0.300-g feed sample is analyzed for its protein content by the modified Kjeldahl method. If 25.0 mL of 0.100 M HCl is required for titration, what is the percent protein content of the sample?A different titration experiment using a 0.127M standardized NaOH solution to titrate a 27.67 mL solution with an unknown Molarity concentration (M) of sulfuric acid (H2SO4) gave the following molarities for 3 trials. Initial Burette Reading (mL) Final Burette Reading (mL) Delivered vol (mL) Acid Concentration (M) Trial 1 0.0358 Trial 2 0.0341 Trial 3 0.0331 From the 3 trials, determine the average Molarity concentration of the H2SO4 to 3 significant digits. Don't include a unit. From the 3 trials, determine the standard deviation of the Molarity concentration of the H2SO4 to 3 significant digits. Don't include a unit.From the 3 trials, determine the relative standard deviation of the Molarity concentration of the H2SO4 to 3 significant digits. Don't include a unit. Do you think the above experiment was accurate? Precise? Explain your answer using supporting values.During titration the following data were collected. A 25 mL poretion of an unknown acid soluation was titrated with 1.0 M NaOH. It required 65mL of the base to neutalize the sample. How many moles of acid are present in 3.0 liters of this unknown solution?
- 4) Students were asked to perform a lab to identify an unknown acid.The experiment has several parts. First the students had to make a sodium hydroxide solution with a molarity of approximately 0.2M. Then the students had to perform a titration with potassium hydrogen phthalate to standardize their solution. Finally they had to perform a titration with the unknown acid to find the pKa and the molar mass, then using theirdata they had to determine which of the weak acids they had used.Part 1) Make a sodium hydroxide solutionA student measured 4.03 grams of NaOH in a tared weigh boat. They then dissolved it into approximately 200 mL of deionizedwater in a beaker. Once it was completely dissolved the student poured the solution into a 500.0 mL volumetric flask. The student rinsed the beaker with deionized water and added the rinse water to the volumetric flask. They repeated with process, adding the rinse water to the volumetric flask again, they mixed this thoroughly. Finally, they…A pharmacist is conducting an analysis of 2.7 g sodium bicarbonate and titrated with 26.05 mL of 0.9987 N HCl. What is the potency or percent assay of sodium bicarbonate? Did it pass the USP requirement? ROUND-OFF TO 2 DECIMAL PLACES. Answer = _% Disposition (Passed or Failed) = _A different titration experiment using a 0.122M standardized NaOH solution to titrate a 26.48 mL solution with an unknown Molarity concentration (M) of sulfuric acid (H2SO4) gave the following molarities for 3 trials. Initial Burette Reading (mL) Final Burette Reading (mL) Delivered vol (mL) Acid Concentration (M) Trial 1 0.0345 Trial 2 0.0334 Trial 3 0.0381 From the 3 trials, determine the average Molarity concentration of the H2SO4 to 3 significant digits. Don't include a unit.
- Use a primary standard to determine an unknown concentration using an acid-base titration. Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory as a primary standard. It has the unwieldy formula of KHC8H404. This is often written in shorthand notation as KHP. If 39.86 mL of a potassium hydroxide solution are needed to neutralize 2.437 grams of KHP, what is the concentration (mol/L) of the potassium hydroxide solution? MA sample containing the amino acid alanine, CH3CH(NH2)COOH, plus inert matter is analyzed by Kjeldahl method. A 2.00 g sample is digested, the ammonia is distilled and collected in 50.0 ml of 0.150 M H2SO4, and a volume of 9.0 ml of 0.100 M NaOH is required for back titration. Calculate the percent alanine in the sample.20 ml of a 100 mg/l phenol solution in water is placed in a vial with 20 ml of a fluorocarbon solvent. The organic phase (fluorocarbon solvent) is immiscible with water. The vial is completely filled with the two phases, thus there is no air or contaminant volatilization. After an hour of agitation, the two phase system reaches equilibrium. The two phases are separated and the fluorocarbon is analyzed for total phenol. The resulting phenol concentration in the organic phase is 16.0 mg/l. Compute the unitless partition coefficient (KF-W) for phenol. Compare your calculation with KF-W obtained from an empirical relationship: log10KF-W 0.71 - 0.862log10 CWsat. KF-W means the partition between the fluorocarbon solvent, F, and water, W. Only typed solution.