Experiment 18 Report Shee Potentiometric Analyse Date A. Molar Concentration of

Question

Experiment 18 Report Shee Potentiometric Analyse Date A. Molar Concentration of a Weak Acid Solution Sample no. ab Sec.Name Desk No. Monoprotic or diprotic acid? Trial I Trial 3 1. Molar concentration of NaOH (mol/L) 2. Volume of weak acid (mL) 3. Buret readng of NaOH, initial (mL) 4. Buret reading NaOH at stoichiometric point, Trial 2 25 final (mL) 5. Volume of NaOH dispensed (mL) 6. Instructor's approval of pH vs. Vsaoe graph 7. Moles of NaOH to stoichiometric point (mol 8. Moles of acid (mol) 9. Molar concentration of acid (mol/L) 10. Average molar concentration of acid (moU/L) B. Molar Mass and the pK, of a Solid Weak Acid Sample no.2 in Trial I Suggested mass Trial 2 , 53 Monoprotic or diprotic acid? Trial 3 1. Mass of dry, solid acid (g) 2. Molar concentration of NaOH (moVL 3. Buret readng of NaOH, initial (mL) 4. Buret reading NaOH at stoichiometric point, 100m final (mL) 5. Volume of NaOH dispensed (mL) 6. Instructor's approval of pH versus Vwon graph 7. Moles of NaOH to stoichiometric point (mol)

Explanation / Answer

part A) To calculate mol of NaOH to the stoichiometric point ,you need molar concentration of NaOH

If molar concentration of NaOH=0.1M=0.1mol/L (taken from part B)

volume of NaOH dispensed =(final burette reading-initial burette reading)=(36ml-0 ml)=36ml

Thus ,mol of NaOH dispensed=molar concentration of NaOH * volume of NaOH dispensed =0.1mol/L *0.036L=0.0036mol

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