Lab 14 Handin done

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Chemistry

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Dec 6, 2023

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Lab 14 Hand-in Name: olivia Landry Purpose: To standardize a sodium hydroxide solution and determine the molarity of an acetic acid solution. Data: Table 1: Standardization of NaOH Molarity of HCl : 0.9535 M Final HCl buret reading (mL) 20.50 mL Final NaOH buret reading (mL) 27.80 mL Initial HCl buret reading (mL) 0.50 mL Initial NaOH buret reading (mL) 0.70 mL Table 2: Titration of unknown HC 2 H 3 O 2 Final HC 2 H 3 O 2 buret reading (mL) 20.30 mL Final NaOH buret reading (mL) 24.80 mL Initial HC 2 H 3 O 2 buret reading (mL) 0.30 mL Initial NaOH buret reading (mL) 0.40 mL Calculations: Part 1: Standardization of NaOH HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) Its balanced 1. Calculate volume of HCl. Convert to L Final HCl buret reading (mL) – Initial HCl buret reading (mL) = mL HCl 20.50 mL - 0.50 mL = 20mL Volume HCl mL x 1 L = L HCl 1000 mL 20 mL * 1/1000ml = 0.02L HCI 2. Calculate volume NaOH used in titration (Table 1). Convert to L Final NaOH buret reading – initial NaOH buret reading = mL NaOH 27.80 mL - 0.70 mL= 27.1mL NaOH Volume NaOH mL x 1 L = L NaOH 1000 mL 27.1mL NaOH*1/1000=0.0271L NaOH

3. Calculate the molarity of NaOH (mol NaOH/L NaOH) Record answer in Table 3. L HCl x mole HCl x mole NaOH x 1 = mol NaOH (M NaOH) L HCl mole HCl L NaOH L NaOH 0.02L HCI * 0.9535 M / 0.02L HCI .* 39.99mol NaOH / 36.46 mol HCI * 1 / 0.0271L NaOH = 38.59 M NaOH Part 2: Titration of unknown HC 2 H 3 O 2 NaOH (aq) + HC 2 H 3 O 2 (aq)  NaC 2 H 3 O 2 (aq) + HOH (l) 4. Calculate volume of HC 2 H 3 O 2 used in titration. Convert to L. Final buret reading HC 2 H 3 O 2 – initial buret reading HC 2 H 3 O 2 = mL HC 2 H 3 O 2 20.30 mL - 0.30 mL = 20 mL HC2H3O2 Volume HC 2 H 3 O 2 in mL x 1 L = L HC 2 H 3 O 2 1000 mL 20 mL HC2H3O2 * 1 /1000 = 0.02L 5. Calculate volume of NaOH used in titration (Table 2). Convert to L. Final buret reading NaOH – initial buret reading NaOH = mL NaOH 24.80 mL - 0.40 mL= 24.40mL NaOH Volume NaOH in mL x 1 L = L NaOH 1000 mL 24.40mL NaOH * 1 / 1000 = 0.0244L 6. Calculate molarity of HC 2 H 3 O 2 (mol HC 2 H 3 O 2 /L HC 2 H 3 O 2 ). Report answer in Table 3. L NaOH x mol NaOH x mol HC 2 H 3 O 2 x ____1_____ = mol HC 2 H 3 O 2 /L HC 2 H 3 O 2 (M HC 2 H 3 O 2 ) from #1 from #2 Coefficients from balanced equation Molarity HCl in Table 1

L NaOH mol NaOH L HC 2 H 3 O 2 0.0244L NaOH * 38.59 / 0.0244L NaOH * 60.00 G/mol / 39.99 g/mol * 1 / 0.02L = 2894 mol HC2H3O2/L HC2H3O2 (M HC2H3O2) Results: Table 3 Part 1 Standardization of Sodium Hydroxide (NaOH) Part 2 Titration of Acetic Acid (HOAc or CH 3 COOH)) Molarity NaOH (mol/L) 38.59 NaOH Molarity HOAc (mol/L) 2894 Conclusion: In the space provided in the hand-in summarize your results (list molarities of NaOH standard and acetic acid). Questions: 1. Vinegar is a solution of acetic acid (the solute) in water (the solvent) with a solution density of 1010 g/L. If vinegar is 0.80 M acetic acid, what is the % by mass concentration of acetic acid in vinegar? Hints: Base your % mass calculation on 1 L of vinegar (solution) You are given the mass of 1 L of vinegar (solution) – density Calculate how many grams of HC 2 H 3 O 2 (solute) are in 1 L (use molarity) % mass = grams solute (acetic acid) x 100 grams solution (vinegar) . .80 * 60.00 G/mol/ 1 mol = 48g 48g / 1010 g/L * 100 = 4.75% Mass 2. How would each of the following errors affect the experimental value of your molarity of acetic? Would your Molarity value be too high, too low, or unaffected? Explain your answer. a. if you titrated your acid sample to a bright pink rather than faint pink endpoint? from #4 from #5 Coefficients from balanced equation Molarity NaOH calculated in #3

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Too high you have add too much in and are unable to use the data. You should stop when after swearing the mixer the light pink color does no disappear b. if your "acetic acid" buret was still wet inside with deionized water when you filled it with acetic acid? Too Low by doing this you have diluted your solution. You should have raised it with the acetic acid after raising with water. c. if you recorded the final volume of NaOH as "39.88 mL" instead of "38.88 mL"? too high you have added more the you were to.