EXPERIMENT: 6 ANALYSIS OF UNKNOWN ACIDS BY ACID/BASE TITRATION (PARTID Materials

Question

EXPERIMENT: 6 ANALYSIS OF UNKNOWN ACIDS BY ACID/BASE TITRATION (PARTID Materials Required: 50 mL buret with clamp and stand phenolphthalein indicator 0.100 M sodium hydroxide (NaOH) unknown sulfuric acid (HSO4) solution 10 mL measuring pipet green pipette pump 11 (for measure pipe) SAFETY PRECAUTIONS: EYE PROTECTION MUST BE WORN AT ALL TIMES IN LABORATORY. IF YOU SPILL ACID OR BASE ON YOU, IMMEDIATELY WASH WITH WATER. NOTIFY YOUR INSTRUCTOR. PROCEED TO THE NEAREST SAFETY SHOWER IF NECESSARY Background: In a titration, the concentration of a substance in a solution is measured through the reaction with a known stoichiometric amount of another reagent. In this experiment, you will obtain a sulfuric acid solution of unknown concentration. To the sulfuric acid solution, you will add a solution of sodium hydroxide of known concentration until all the acid is exactly neutralized. An acid-base indicator (phenolphthalein) will change color when the acid is exactly neutralized, signaling the equivalence point. The chemical equation for this neutralization reaction is as follows: Haso, (aq) 2NaOH (aq)- Na2SO4 (aq) 2H20 a) + + From the measured volume of sodium hydroxide needed to reach the equivalence point, the concentration of the acid solution can be determined using the following mathematical equation: Where a and b are stoichiometric coefficients of the acid and base, respectively. VA and Va are the volumes of the acid and base used in the titration and Ma is concentration of the base. 37

Explanation / Answer

Volume of Acid (V1) = 20 mL

volume of Base (V2)= 15.7 - 1 = 14.7 mL

molarity of base (M2)= 1 M (assume)

molarity of Acid (M1)= ?

H2SO4 + 2 NaOH -----> Na2SO4 + H2O

we have the formula M1V1/n1 = M2V2/n2 and n1 = 1 and n2 = 2

M1 * 20/1 = 1 * 14.7/2

M1 = molarity of acid = 0.3675 M

Volume of Acid (V1) = 20 mL

volume of Base (V2)= 38.1 - 15.7 = 22.4 mL

molarity of base (M2)= 1 M (assume)

molarity of Acid (M1)= ?

H2SO4 + 2 NaOH -----> Na2SO4 + H2O

we have the formula M1V1/n1 = M2V2/n2 and n1 = 1 and n2 = 2

M1 * 20/1 = 1 * 22.4/2

M1 = molarity of acid = 0.56 M

Average concentration of sulfuric acid = (0.3675 + 0.56)/2 = 0.4637 M

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