124 Report Sheet Titration of Acids and Bases JAverage percent of KHP Standard d

Question

124 Report Sheet Titration of Acids and Bases JAverage percent of KHP Standard deviation- Calculations of percent KHP and standard deviation (show using equations with units): QUESTIONS 1. Write the balanced chemical equation for the reaction of KHP with NaOH. 2. Suppose your laboratory instructor inadvertently gave you a sample of KHP contaminated with NacI to use in standardizing your NaOH. How would this affect the molarity you calculated for your NaOH solu- tion? Justify your answer , 3. How many grams of NaOH are needed to prepare 500 ml of 0.125 MNaOH? 2,S 4. A solution of malonic acid, HC,H 04, was standardized by t with 0.1000 M NaOH solution. If NaOH solution is required to neutralize completely 12.95 ml of the malonic acid solu- tion, what is the molarity of the malonic acid solution? 129S Dx 0.6207 00307

Explanation / Answer

Ans. # Note: The average % KHP and SD can’t be calculated unless the data for two or more trials are provided.

#1. Balanced reaction:

                        KHC8H4O4(aq) + NaOH(aq) -------> Na2C8H4O4(aq) + H2O(l)

#2. NaCl, being a salt of strong acid (HCl) and strong base (NaOH) does not alter the pH of the solution – both the ions Na+ and Cl- simply act as spectator ion.

So, titration of a NaCl-contaminated KHP sample with NaOH does not affect the titration endpoint. Since titration endpoint remains unaffected of NaCl-contamination, the calculated molarity of NaOH solution also remains unaffected.

#3. Required moles of NaOH = Molarity x Volume of solution in liters

                                                = 0.125 M x 0.500 L

                                                = 0.0625 mol

Required mass of NaOH = Moles x MW = 0.0625 mol x (40.0 g/ mol) = 2.500 g

#4. At equivalence point, the total number of moles of H+ from acid is equal to total number of OH- from base.

That is-           x(M1V1), acid = y(M2V2), base             - equation 1

            Where, x = moles of H+ produced per mol acid = 2 for H2C3H2O4

                        y = moles of OH- produced per mol base = 1 for NaOH

                        V and M are volume and molarity of respective solution.

# Putting the values in equation 1-

            2 x (M1 x 12.95 mL) = 1 x (0.1000 x 20.76 mL)

            Or, M1 = (0.1000 x 20.76 mL) / (2 x 12.95 mL)

            Hence, M1 = 0.08015

Therefore, molarity of H2C3H2O4 solution = 0.08015 M

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.