(a) How many milliliters of 0.165 M HCl are needed to neutralize completely 35.0
ID: 501603 • Letter: #
Question
(a) How many milliliters of 0.165 M HCl are needed to neutralize completely 35.0 mL of 0.101 M Ba(OH)2 solution? ______ml
(b) How many milliliters of 3.50 M H2SO4 are needed to neutralize 75.0 g of NaOH?
_______mL
(c) If 55.8 mL of BaCl2 solution is needed to precipitate all the sulfate in a 544 mg sample of Na2SO4 (forming BaSO4), what is the molarity of the solution?
________M
(d) If 47.5 mL of 0.375 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)
a)
2 HCl + Ba(OH)2 ------------> BaCl2 + 2 H2O
M1 V1 / n1 = M2 V2 / n2
0.165 x V1 / 2 = 0.101 x 35 / 1
V1 = 42.85
volume of HCl needed = 42.8 mL
(b)
H2SO4 + 2 NaOH ------------> Na2SO4 + 2 H2O
moles of NaOH = 75 / 40 = 1.875 mol
moles of H2SO4 = 1.875 / 2 = 0.9375
molarity = moles / volume
3.50 = 0.9375 / volume
volume of H2SO4 = 0.268 L
volume of H2SO4 = 268 mL
(c)
moles of Na2SO4 = 0.544 / 142 = 3.83 x 10^-3 mol
moles of BaCl2 = moles of Na2SO4
moles of BaCl2 = 3.83 x 10^-3 mol
molarity = 3.83 x 10^-3 / 0.0558
Molarity = 0.0687 M
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