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0.02 M NaOH reagent was standardized using 0.1000 g Potassium hydrogen phthalate (204.22 g/mol) and it required 25.8 mL to reach end point.
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- 0.02 M NaOH reagent was standardized using 0.1000 g Potassium hydrogen phthalate (204.22 g/mol) and it required 25.8 mL to reach end point. What is the molar concentration required?A 0.5027 g of KHP (KC8H5O4; MW = 204.22 g/mol) required 11.90 mL of NaOH titrant to reach the phenolphthalein endpoint. Calculate the standardized concentration of NaOH.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 acidA 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 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.
- Calculate the percentage CH3COOH in a sample of vinegar from the following data.Sample = 15.00 g, NaOH used = 43.00 ml ; 0.600 N H2SO4 used for back titration =0.250 ml ; 1.00ml NaOH is equivalent to 0.0315 g H2C2O4. 2H2O.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 acidA solution of HCl was titrated against sodium carbonate. What is the titer for Trial 1 in the given data? T2 0.3562 0.3479 0.3042 Weight (g) Initial V (ml) 0.80 1.60 35.20 36.70 39.80 Final V (ml) Vol HCI used (ml) 35.10 N of HC (eq/L) 0.1954 0.1870 Average N of HCl (eq/L) A
- Questions 11-14 refer to the same strong base/strong acid (SB/SA) titration. A 15.00 mL solution of 0.100 M sodium hydroxide (NaOH) is being titrated with 0.250 M hydrobromic acid (HBr). What is the total solution volume (in mL) at the equivalence point? (Nearest whole number) Type your answer...What is the percentage of total acid expressed as acetic acid in a sample of vinegar if 4.0g of vinegar requires 29.5ml of 0.153N KOH for an endpoint with phenolphthalein indicator? (molar mass of acetic acid: HC2H3O2 = 60.05g/mol a. 3.995% b. 6.776% c. 8.650% d. 12.68%Part A: Standardization of a Sodium Hydroxide Solution Titration 1 Titration 2 Titration 3 Mass of 125 mL flask 45.849g 46.715g 44.953g Mass of flask and KHP 46.849g 47.745g 46.003g Initial buret reading (mL) 0.5 ml 0.5 ml 0.5 ml Final buret reading (mL) 27.8 ml 26.5 ml 26.7 ml Volume of NaOH used (mL) 45.11 ml 45.06 ml 45.14 ml Calculations Titration 1 Titration 2 Titration 3 Moles of KHP Moles of NaOH Molarity of NaOH Average Molarity of NaOH: _______________