If you were to filter and dry the barium sulfate precipitate that formed in the
ID: 511400 • Letter: I
Question
If you were to filter and dry the barium sulfate precipitate that formed in the titration in part A, what mass would you expect to obtain based on your molarity of the barium hydroxide solution and the volume used in trial 2? DATA SHEET Part A-Analysis of Unknown Barium Hydroxide Solution. ID Trial 1 Volume of Ba (OH)2(aq) solution dispensed 10. oomt Molarity of Hzso4(aq) solution used Total volume of H2Soalaq) Conductivity Dis (mL) KAS/cm) Oo 152.5 Approximate value of Equivalence Point Som Approximate Molarity of Ba (OH)2(aq) solution Show how you calculated the approximate molarity of the Ba(OH)2(aq) solution: malar equation m A SayExplanation / Answer
1. 1 mol Ba(OH)2 reacts with 1 mol H2SO4 and produces 1 mol BaSO4 as a precipitate. Therefore how many moles of Ba(OH)2 is reacted = how many moles of BaSO4 produced.
2. First we will find out how many moles present in 10 mL of 0.056 M solution of Ba(OH)2
moles = 10 mL x 10-3 L x 0.056 M = 0.00056 mol
3. Covert moles to mass by multiplying with molar mass of barium sulfate
mass of barium sulfate produced as a precipitate = 0.00056 mol x 233.38 g/mol = 0.1306928
mass of barium sulfate produced as a precipitate = 0.13 grams
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