(a) How many milliliters of 0.145 M HCl are needed to neutralize completely 45.0
ID: 902503 • Letter: #
Question
(a) How many milliliters of 0.145 M HCl are needed to neutralize completely 45.0 mL of 0.101 M Ba(OH)2 solution? ________ml
(b) How many milliliters of 2.50 M H2SO4 are needed to neutralize 25.0 g of NaOH?_______ mL
(c) If 56.8 mL of BaCl2 solution is needed to precipitate all the sulfate in a 534 mg sample of Na2SO4 (forming BaSO4), what is the molarity of the solution? ________M
(d) If 37.5 mL of 0.125 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solution? ______g
Explanation / Answer
(a) moles Ba(OH)2 = 0.101 M x 45 ml = 4.545 mmol
Volume HCl needed = 2 x 4.545 mmol/0.145 M = 62.70 ml
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(b) moles NaOH = 25 g/40 g/mol = 0.625 mol
Volume H2SO4 needed = 0.625 mol x 1000/2 x 2.50 M = 125 ml
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(c) moles Na2SO4 = 534 mg/142.04 g/mol = 3.76 mmol
molarity of BaCl2 solution = 3.76 mmol/56.8 ml = 0.0662 M
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(d) moles HCl used = 0.125 M x 37.5 ml = 4.7 mmol
grams of Ca(OH)2 = 4.7 x 10^-3 mol x 74.093 g/mol/2 = 0.174 g
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