A 20.00 mL wine sample is titrated with a 1.04×10-1M solution of sodium hydroxide. The equivalence point is determined to be 25.926 mL. What is the acidity of the wine expressed as (g tartaric acid / L; MW tartaric acid = 150.085 g/mole)?
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A 20.00 mL wine sample is titrated with a 1.04×10-1M solution of sodium hydroxide. The equivalence...
A 40.00 mL wine sample is titrated with a 1.04 times 10^-1M solution of sodium hydroxide. The equivalence point is determined to be 29.267 mL. What is the acidity of the wine expressed as (g tartaric acid/L, MW tartaric acid 3150.085 g/mole)? 5.71 11.4 11.42 3.38 5.710 The acidity of a wine sample is determined to be 14.21 g tartaric acid/L What is the mass of tartaric acid needed to prepare an equivalent concentration solution in a 50 mL volumetric...
The acidity of a wine sample is 1.05×101, expressed as (g citric acid / L wine; MW citric acid = 192.124 g/mole) . The sample is titrated with a 0.118 M solution of sodium hydroxide. The equivalence point is determined to be 39.740 mL. What is the volume of the wine sample analyzed in mL?
The acidity of a wine sample is determined to be 13.08 g tartaric acid/L. What is the mass of tartaric acid needed to prepare an equivalent concentration solution in a 50 mL volumetric flask? (MW tartaric acid = 150.085 g/mole.)
A 20.00-mL solution of 0.120 M nitrous acid (Ka = 4.0 × 10–4) is titrated with a 0.215 M solution of sodium hydroxide as the titrant. What is the pH of the acid solution at the equivalence point of titration? (if needed: Kw = 1.00 × 10–14)
Part A When titrated with a 0.1198 M solution of sodium hydroxide, a 58.00 mL solution of an unknown polyprotic acid required 20.15 mL to reach the first equivalence point. Calculate the molar concentration of the unknown acid. O A¢ * R O ? Submit Request Answer Part B The titration curve was found to have three equivalence points. What volume of the sodium hydroxide solution was necessary to fully titrate the unknown acid solution? IVO ADD A O O...
Part A When titrated with a 0.1038 M solution of sodium hydroxide, a 52.00 mL solution of an unknown polyprotic acid required 22.86 mL to reach the first equivalence point. Calculate the molar concentration of the unknown acid. VALDO ? Submit Request Answer
A 20.00 mL sample of V2+ is titrated to the equivalence point with 23.71 mL of 0.0646 M KMnO4in a solution with a pH = 1.11. Calculate the potential of the solution. The reaction of the sample is V2+ -> V3+. The titrant reaction is the one that takes MnO4- to Mn2+. Use Appendix 13. Express the answer with 2 decimal places.
A 22.55-mL sample of hydrochloric acid solution requires 20.00 mL of 0.148 M sodium hydroxide for complete neutralization. What is the concentration of the original hydrochloric acid solution? 1 м
When a 23.2 mL sample of a 0.345 M aqueous acetic acid solution is titrated with a 0.336 M aqueous sodium hydroxide solution, what is the pH after 35.7 mL of sodium hydroxide have been added? pH = What is the pH at the equivalence point in the titration of a 27.6 mL sample of a 0.423 M aqueous hydrocyanic acid solution with a 0.431 M aqueous potassium hydroxide solution? pH=
A 50.0 mL sample of 0.150 M sodium hydroxide is titrated with 0.250 M nitric acid. Calculate: a. the pH after adding 10.00 mL of HNO3 b. the pH after adding 40.00 mL of HNO3 c. the volume required to reach the equivalence point d. the pH at the equivalence point