A 0.5220 −g sample of an unknown monoprotic acid was titrated with 9.94×10−2 M NaOH. The equivalence point of the titration occurs at 23.86 mL . Determine the molar mass of the unknown acid.
A 0.5220 −g sample of an unknown monoprotic acid was titrated with 9.94×10−2 M NaOH. The...
A 0.5216 ?g sample of an unknown monoprotic acid was titrated with 9.94×10?2 M NaOH. The equivalence point of the titration occurs at 23.76 mL . Part A Determine the molar mass of the unknown acid. Express your answer using three significant figures
A 0.5240 −g sample of an unknown monoprotic acid was titrated with 9.94×10−2 M NaOH. The equivalence point of the titration occurs at 23.90 mL . Determine the molar mass of the unknown acid. Use the Ksp values to calculate the molar solubility of each of the following compounds in pure water. PART A---- MX (Ksp = 8.57×10−36)Express your answer in moles per liter. PART B----PbCl2 (Ksp = 1.17×10−5) Express your answer in moles per liter. PART C-----Ni(OH)2 (Ksp =...
A 0.4352-g sample of an unknown monoprotic acid is dissolved in water and titrated with standardized potassium hydroxide. The equivalence point in the titration is reached after the addition of 31.14 mL of 0.1833 M potassium hydroxide to the sample of the unknown acid. Calculate the molar mass of the acid. _____ g/mol
A sample of 0.2140 g of an unknown monoprotic weak acid was dissolved in 25.0 mL of water and titrated with 0.0950 M NaOH. The acid required 15.50 mL of NaOH to reach the equivalence point. What is the molar mass of the unknown acid?
A sample of 0.2140 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950 M NaOH. The titration required 30.0 mL of base to reach the equivalence point, at which point the pH was 8.68. a) What is the molecular weight of the acid? b) What is the pKa of the acid?
1) A solution of a weak monoprotic acid of unknown concentration was titrated with 0.23 M NaOH. If a 100.-mL sample of the acid solution required exactly 10. mL of the NaOH solution to reach the equivalence point, what was the original concentration of the weak acid? 2) During the titration on problem (2B), after 5.0 mL of NaOH addition, the pH = 3.68. What is the Ka of the weak acid? please show steps i have an exam tomorrow
A 0.446-g sample of an unknown monoprotic acid was titrated with 0.105 M KOH and the resulting titration curve is shown in the figure(Figure 1). Part A Determine the molar mass of the acid. Part B Determine pKa of the acid.
A 0.699 g sample of a monoprotic acid is dissolved in water and titrated with 0.240 M NaOH. What is the molar mass of the acid if 30.0 mL of the NaOH solution is required to neutralize the sample? molar mass: g/mol
A 0.173 g sample of a monoprotic acid is dissolved in water and titrated with 0.120 M NaOH. What is the molar mass of the acid if 10.5 mL of the NaOH solution is required to neutralize the sample? molar mass: 274.6 molar mass: 274.6 g/mol g/mol
A sample of 0.2140 grams of an unknown monoprotic weak acid was dissolved in 25.0mL of water and titrated with 0.0950M NaOH. The acid required 15.50mL of NaOH to reach the equivalence point. What is the molar mass of the unknown acid?