A 0.5240 −g sample of an unknown monoprotic acid was titrated with 9.94×10−2 M NaOH. The...

Question

Question

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 = 5.48×10⁻¹⁶) Express your answer in moles per liter.

Answer 1

Answer #1

Answer 2

Similar Homework Help Questions

A 0.5216 ?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 0.5220 −g sample of an unknown monoprotic acid was titrated with 9.94×10−2 M NaOH. The...

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.
Use the Ksp values to calculate the molar solubility of each of the following compounds in...

Use the Ksp values to calculate the molar solubility of each of the following compounds in pure water. Part A MX ( K sp = 5.30×10−11) Express your answer in moles per liter. Part B Ag2CrO4 (Ksp = 1.12×10−12) Express your answer in moles per liter. Part C Ni(OH)2 (Ksp = 5.48×10−16) Express your answer in moles per liter.

A 0.4352-g sample of an unknown monoprotic acid is dissolved in water and titrated with standardized...

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 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).

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.
II Review | Constants | Periodic Table Use the Ksp values to calculate the molar solubility of each of the following co...

II Review | Constants | Periodic Table Use the Ksp values to calculate the molar solubility of each of the following compounds in pure water. Part A You may want to reference (Pages 769 - 775) Section 17.5 while completing this problem. MX (Ksp = 2.37x10-36) Express your answer in moles per liter. EVO AQ R o 2 ? S = M Submit Request Answer Part B PbCl2 (Ksp = 1.17x10-5) Express your answer in moles per liter. V ALO...

5. The Ka and Molar Mass of a Monoprotic Weak Acid a. Suppose that–unknown to you–the...

5. The Ka and Molar Mass of a Monoprotic Weak Acid a. Suppose that–unknown to you–the primary standard KHP (potassium hydrogen phthalate, KHC8H4O4) had a potassium iodide impurity of approximately one percent by mass. How would this have influenced the calculated molarity of your sodium hydroxide solution? Would your calculated value be too low, too high, or unchanged? Explain your answer. b. Sketch a typical titration curve for a monoprotic weak acid titrated with a strong base. Label the axes...
A sample of 0.2140 g of an unknown monoprotic weak acid was dissolved in 25.0 mL...

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...

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...

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