Experiment Eleven Preparation and Standardization of a Sodium Hydroxide Solution

Question

Experiment Eleven Preparation and Standardization of a Sodium Hydroxide Solution Name Data Sheet Laboratory Section PART B. STANDARDIZATION OF NAOH tl 1. Weight of paper + KHP 012 g Lad74g Dakeg 2. Weight of Paper 3. Weight of KHP (#1-#2) 010797 g 0,02% 0:07 g 4, Number of EW of KHP (#3/2042) 5. Final buret reading 6. Initial buret reading 7a. Volume of NaOH used (#5_#6) 2.26 mL 344 mL S01 mL 7b. Volume of NaOH used (7a/1000) 04120, 04361; Orq o.oo34 BQ ml O.03 ml. 322ml had Dot&L; 0"0xgL0myL doo, 7 000,- ((In do trON 1 8. Normality of NaOH solution Calculation -wt. of KHP/204.2 -7Liters of NaOH #8W KHP-#EWKHP (#4)/(#7b) Use equation (12): NB Liters of NaOH 01127 9. Average normality of NaOH solution 75

Explanation / Answer

NB = weight of KHP/204.2/Liters of NaOH used = EW KH/Liters of NaOH used= EW KH(#/4)(#7b)

KC8H4O4H + NaOH = KC8H4O4Na + H2O

1 mol of KHP neutralizes 1 mol of NaOH

For trial1:

Weight of KHP = 0.0797 g

Molar mass of KHP =204 g/mol

Moles of KHP = 0.0797/204 g/mol = 0.00039 moles

Moles of NaOH = Moles of KHP =0.00039 moles

Molarity of NaOH = Moles of NaOH/ Liters of NaOH

Molarity of NaOH = Normality of NaOH (NaOH has 1 OH)

NB = 0.00039/0.00226 = 0.173 N

For trial2:

NB = 0.00037/0.00364 = 0.102N

For trial3:

NB = 0.00035/0.00308 = 0.102N = 0.114N

Average normality of NaOH solution = (0.173 + 0.102 + 0.114)/3 N = 0.130 N

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

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