Experiment 18 Report Sh Potentiometric Analys Desk No. lob Sec. Name A. Molar Co

Question

Experiment 18 Report Sh Potentiometric Analys Desk No. lob Sec. Name A. Molar Concentration of a Weak Acid Solution Sample no. A Monoprotic or diprotic acid? cotic Trial I Trial 2 Trial 3 1. Molar concentration of NaOH (moV/L) 2. Volume of weak acid (mL) 25 m Buret readng of NaOH, initial (mL.) Cine-? 3. 4. Buret reading NaOH at stoichiometric point ML final (mL) 5. Volume of NaOH dispensed (mL) 6. Instructor's approval of pH vs. VAson graph 7. Moles of NaOH to stoichiometric point (mo) 8. Moles of acid (mol) 9. Molar concentration of acid (moUL) 10. Average molar concentration of acid (molU/L) B. Molar Massand the pK, of a Solid Weak Acid Sample no. .Monoprotic or diprotic acid? ini pcohc Suggested mass Trial 1 Trial 2 5 3 1. Mass of dry, solid acid (g) 2. Molar concentration of NaOH (molL) 3. Buret readng of NaOH, initial (mL.) 4. Buret reading NaOH at stoichiometric point, nL 74 ML final (ml.) 5. Volume of NaOH dispensed (mL) 6. Instructor's approval of pH versus VNon graph 7. Moles of NaOH to stoichiometric point (mol) S. Moles of acid (mol) . Molar mass of acid (g/mol)

Explanation / Answer

For trial 1 you used NaOH with a concentration of 0.1 mol / L

you used 36 ml or 0.036 Liters (at the stoichiometric point)

moles = molarity * volume

moles = 0.036 * 0.1 = 0.0036 moles of naoh

since you have a monoprotic acid then 1 mole of naoh requires 1 mole of acid so you have 0.0036 moles of acid

to calculate the molarity you need the volume (25 ml from the statement) so

molarity = moles / volume

molarity = 0.0036 / 0.025 = 0.144 M

For trial 2 apply the same thing

moles of naoh = 0.1 * 0.038 = 0.0038 moles of NaOH

1 mole of naoh neutralizes 1 mole of monoprotic acid so you have 0.0038 moles of monoprotic acid

molarity = moles / volume = 0.0038 / 0.025 = 0.152 M

average (0.152 , 0.144) = 0.148 M

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