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thanks Question 23 (2.5 points) Saved A 25.00-ml sample of propanoic acid, CH3CH2COOH, of unknown concentration...
QUESTION 1 A 25.00-ml sample of propionic acid. HC H 50 of unknown concentration was titrated with 0.151 M KOH. The equivalence point was reached when 41.28 ml of base had been added. What is the hydroxide ion concentration at the equivalence point? K, for propionic acid is 13 x 10 at 25°C. O A 1.1 x 10 M 93-8.5 x 10 M OC 1.5x 10M OD. 1.0 x 10PM O E 1.1 X 10M
Lone Star CHEM 412 EXamNan - 9.A 25.00-mL sample of propionic acid, HC3HO2, of unknown concentration was titrated with 0.183 M KOH The equivalence point was reached when 41.42 mL of base had been added. What is the hydroxide-ion concentration at the equivalence point? K, for propionic acid is 1.3 x 10-5 at 25°C. a. 1.0x 10-7 M b. 1.2 x 103 M c. 1.1 x 105 M Da S C d. 9.4 x 10 M (e>25Erehle y e. 1.5...
1. 50.00 mL of 0.1000 M propanoic acid (CH3CH2COOH – Ka = 1.34 X 10-5) is titrated with 0.2000 M KOH. Calculate the pH at the following points in the titration: 1) Initial pH – no KOH has been added. 2) 5.00 mL of KOH has been added. 3) 12.50 mL of KOH has been added. 4) At the equivalence point. (Calculate the volume of KOH to reach the equivalence point & identify a good indicator.) 5) Provide a sketch...
a) A 25.00-mL sample of monoprotic acid was titrated with 0.0800 M potassium hydroxide solution. The equivalence point was reached after 18.75 mL of base was added. Calculate the concentration of the acid. b) A 15.00-mL sample of 0.120 M nitric acid was titrated with 0.0800 M potassium hydroxide. Calculate the pH of the sample when 10.00 mL of the base has been added.
JUL &coursejd_40296_1&new_attempt=1&content_ide_1144581.18 Question Completion Status: 35.0 mL of 0.10M propanoic acid CH3CH2COOH (K 1.3 x 10") is added to a 100.0 mL volumetric flask. Also added to the flask is 25.0 mL of 0.10M NaOH and water to fill the flask to the 100.0 ml line. How many moles of conjugate base would be formed? QUESTION 11 35.0 mL of 0.10M propanoic acid CH3CH2COOH (Ka - 1.3 x 10 ) is added to a 100.0 ml volumetric flask. Also added...
ration 46. A 30.0-ml sample of 0.165 M propanoic acid is titrated with 0.300 M KOH. Calculate the pH at each volume of added base: OmL, 5 mL, 10 mL, equivalence point, one-half equivalence point. 20 mL, 25 mL. Use your calculations to make a sketch of the titration curve. ed by
68.5 mL of a HNO3 solution with unknown concentration is titrated with 0.150 M KOH solution. The end point is reached after 25 m/L of KOH solution is added. what is the molarity of the HNO3 solution ? 1 р . HNO3 + KOH --> KNO3 + H20 68.5 mL of a HNO3 solution with unknown concentration is titrated with 0.150 M KOH solution. The endpoint is reached after 25.0 mL of KOH solution is added. What is the molarity...
What is the concentration of 25.00 mL of an unknown monoprotic acid if 18.24 mL of standardized 0.125 M sodium hydroxide solution was required to reach the equivalence point of the titration?
0.1945 M Question 14 0/1 point 25.00 ml of an unknown triprotic acid solution was titrated by 0.1729 M KOH. The third end point was observed when 44.55 mL KOH was delivered. What was the concentration of the unknown acid in the initial solution? 0.09180 M • 0.3081 M 0.1027 M 0.2628 M 0.03423 M
Introductory Chemis Same 2018 "Name: 1. A 25.00-ml sample of an H SO, solution of unknown concentration is titrated with a 0.1322 M KOH solution. A volume of 41.22 ml. of KOH is required to reach the equivalence point. What is the concentration of the unknown H SO, solution? 2. What volume in milliliters of a 0.121 M sodium hydroxide solution is required to reach the equivalence point in the complete titration of a 10.0-ml. sample of 0.102 M sulfuric...