Iaad bat A. Molar tration of a Weak Acid Solution SanplearCanet n acid Monpofic

Question

Iaad bat A. Molar tration of a Weak Acid Solution SanplearCanet n acid Monpofic Monoprotic or diprotic acid? Trial 1 Trial 2 Trial 3 1. Molar concentration of NaOH (moU/L) 2. Volume of weak acid (mL) 3. Buret readng of NaOH, initial (mL) 4, Buret reayng NaOH at s oichiometric point, 51mL Lo 2 ml -7.5mL final (mL 5. Volume of NaOH dispensed (mL) 6. Instructor's approval of pH vs. Vaoe graph 7. Moles of NaOH to stoichiometric point (mol) 8. Moles of acid (mol) 9. Molar concentration of acid (moVL) 10. Average molar concentration of acid (moVL) 5.0x10 8 5.35x10-3 0.23M- B. Molar Mass and the pK of a Solid Weak Sample no. pocl 05 Monoprotic ordiprotic acid Suggested mass Trial 3 Trial 2 05 0 Trial I 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, final (mL) 5. Volume of NaOH dispensed (mL) 6. Instructor's approval of pH versus V-H graph 7. Moles of NaOH to stoichiometric point (mol) 8. Moles of acid (mol) 9. Molar mass of acid (g/mol) 10. Average molar mass of acid (g/mol) 11. Volume of NaOH halfway to stoichiometric point (mL) 12. pk, of weak acid (from graph) 13. Average p Show calculations for Trial 1 on the next page. Experiment 18 235 141

Explanation / Answer

B. For the second part

diprotic acid

Trial 1

7. moles NaOH = 0.1 M x 9 ml = 0.9 mmol

8. moles acid = 0.9/2 = 0.45 mmol

9. molar mass acid = 0.055 g/0.45 x 10^-3 mol = 122.2 g/mol

Same for Trial 2,

9. molar mass acid = 0.051 g/(0.1 M x 10.5 ml/2) x 10^-3 = 97.14 g/mol

10. average molar mass acid = 110 g/mol

graph cannot be done with these points. We need more points.  

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