Crystals can be weighed directly into a 125-mL Erlenmeyer flask or transferred i

Question

Crystals can be weighed directly into a 125-mL Erlenmeyer flask or transferred into a 250-mL flask from weighing paper or a weighing boat. (Tare it to zero.) 9. Add about 20 mL of water to the flask and add three drops of phenolphthalein indicator Use techniques demonstrated by your instructor to titrate the acid to the endpoint, the point at which the solution just chan burette beyond the bottom line (25 ml ges pink and stays pink for at least 30 seconds. DO NOT DRAIN the r 50 mL). Rinse the sides of the flask with water buret. (Note the direction of the scale.) Pink solutions may be poured down the drain 11. Rinse the Erlenmeyer flask and repeat steps 8-9 two more times (three trials total). If the burette is still more than half full, the next titration can start where the last one ended. Otherwise refill the burette with NaOH solution. If the first titration required less than 10 ml of base, use enough acid in subsequent titrations so that they require at least 10 mL to reach the end point. 12. Unless you will go right on to titrate vinegar, drain excess NaOH solution back into the storage bottle, seal the bottle tightly and store it for the next lab period. 13. Thoroughly rinse the burette, including the tip, with tap water and drain it completely before returning it. C. Data for Standardization: Table I Trial 1 Trial 2 Trial 3 0.50 0.50504 0.5046 5.95m 7 Mass KHP (s) Final burette readingmL i3 40mL Initial burette readingml ntial burtte redingL .1mL 5mL Volume of base used (subtract initial bom final) 26mL3. mL 13.95 mL Write a molecular equation for the reaction above Write the equation as a net ionic equation (note that HCsHO is a weak acid.)

Explanation / Answer

C) Standarization of base

Data

D) molecular equation (let KOH be the base)

KC8H5O4 + KOH ---> K2C8H4O4 + H2O

Net ionic equation

H+ + OH- ---> H2O

Molarity of base

Trial 1 = 0.5041 g/204.2 g/mol x 0.01267 L = 0.195 M

Trial 2 = 0.5050 g/204.2 g/mol x 0.01328 L = 0.186 M

Trial 3 = 0.5046 g/204.2 g/mol x 0.01325 L = 0.186 M

Average = 0.19 M

Part II : Determination of acid molarity

4. Molarity of vinegar

Trial 1 = 0.19 M x 23.72 ml/5 ml = 0.90 M

Trial 2 = 0.19 M x 24 ml/5 ml = 0.91 M

Trial 3 = 0.19 M x 23.96 ml/5 ml = 0.91 M

Average = 0.906 M

5. % acetic acid = grams of acetic acid/grams of vinegar solution

mass of vinegar solution = 5 ml x 1 g/ml = 5 g

mass of acetic acid = 0.906 M x 5 ml x 60.05 g/mol/1000 = 0.272 g

So, % acetic acid in vinegar solution = (0.272/5) x 100 = 5.44%

B. Determination of molarity of unknown acid

7. Molarity of acid

Trial 1 = 0.19 M x 22.51 ml/10 ml = 0.427 M

Trial 2 = 0.19 M x 22.25 ml/10 ml = 0.423 M

Trial 3 = 0.19 M x 22.42 ml/10 ml = 0.426 M

8. Average molarity of unknown acid = 0.425 M

10. ml of KOH required for neutralization of H2SO4 = 0.6552 M x 25 ml/0.8916 M = 18.37 ml

12. moles of Ca(OH)2 = 4.68 g/74.093 g/mol = 0.063 mol

2 moles of HNO3 needed to neutralize 1 mole of Ca(OH)2

Volume of HNO3 required = 0.063 mol x 2/0.3770 M = 334.22 ml

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