1. 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 0.8053 g Volume of sample NaOH (mL) used KHP Weight (g) Purity NaOH (mL) used Determine the following: 50.00 mL 99.80% 33.20 mL 40.60 mL a. Molarity of NaOH b. % (w/v) acetic acid

1. 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 0.8053 g Volume of sample NaOH (mL) used KHP Weight (g) Purity NaOH (mL) used Determine the following: 50.00 mL 99.80% 33.20 mL 40.60 mL a. Molarity of NaOH b. % (w/v) acetic acid

Chemistry & Chemical Reactivity

10th Edition

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter17: Principles Of Chemical Reactivity: Other Aspects Of Aqueous Equilibria

Section: Chapter Questions

Problem 54PS: If 55 mg of lead(II) sulfate is placed in 250 mL of pure water, does all of it dissolve? If not, how...

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If 55 mg of lead(II) sulfate is placed in 250 mL of pure water, does all of it dissolve? If not, how much dissolves?
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 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. 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
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 acid
. A 4.59 mL sample of HCl, specific gravity 1.3, required 50.5 mL of 0.9544N NaOH in a titration. Calculate the % w/w HCl.
A 250.0-mg sample of an organic weak acid was dissolved in an appropriate solvent and titrated with 0.0556 M NaOH, requiring 32.58 mL to reach the end point. Determine the compound’s equivalent weight.
What is a primary reason for needing to standardize strong base titrants like NaOH and KOH? 1)Standardization is the process of dissolving NaOH or KOH in aqueous solution.2)NaOH and KOH are both inexpensive, highly soluble, and easy-to-use bases.3)Standardization is the process of reacting a strong base like NaOH or KOH with an acid.4)NaOH and KOH both react with moisture in the air thereby making accurately-determined concentrations of their solutions difficult.5)NaOH and KOH are not very soluble in aqueous solutions making the concentrations difficult to determine.
1a) A 0.605 gm sample of Sodium Carbonate and Sodium bicarbonate , the solution titrated with 0.106N HCI , 40.2 ml needed from the titrant to reach phenolphthalein , and when titrated in presence of blue bromophenol found the total acid volume is 95.4 ml . calculate the % for the constituents ?
Three drops of the phenolphthalein chemical indicator should be added to the acidic analyte sample (KHP or citric acid) before every titration. You will then carefully add the base to the analyte until the solution turns O a. from clear to a very pale pink color. O b. from a pale pink color to totally clear. O. from a dark pink color to totally clear. O d. from clear to a very dark pink color.
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