(a) How many milliliters of 0.165 M HCl are needed to neutralize completely 45.0
ID: 949904 • Letter: #
Question
(a) How many milliliters of 0.165 M HCl are needed to neutralize completely 45.0 mL of 0.101 M Ba(OH)2 solution?
________ml
(b) How many milliliters of 1.50 M H2SO4 are needed to neutralize 75.0 g of NaOH?
________ mL
(c) If 54.8 mL of BaCl2 solution is needed to precipitate all the sulfate in a 554 mg sample of Na2SO4 (forming BaSO4), what is the molarity of the solution?
_________ M
(d) If 27.5 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) How many milliliters of 0.165 M HCl are needed to neutralize completely 45.0 mL of 0.101 M Ba(OH)2 solution?
________ml
ratio is 2:1 so
mol of Ba(OH)2 = MV = 45*0.101 = 4.545
mol of HCl = 2*¨mol of base = 4.545*2 = 9.09
V = mol/M = 9.09/(0.165) = 55.090ml
(b) How many milliliters of 1.50 M H2SO4 are needed to neutralize 75.0 g of NaOH?
________ mL
mol of NAOH= mass/MW = 70/40 = 1.875
1 mol of acid neutralize 2 mol of base
then we need1.875/2 = 0.9375 mol of acid
V = mol/M = 0.9375/1.5 = 0.625 L = 625 mL
(c) If 54.8 mL of BaCl2 solution is needed to precipitate all the sulfate in a 554 mg sample of Na2SO4 (forming BaSO4), what is the molarity of the solution?
_________ M
1 mol of BaCl2 need 1 mol of NA"SO4 to precipitate
mol of Na2SO4 = mass/MW = 554*10^-3)/142.04 = 0.00390030977 mol
then we have = 0.00390030977mol of BACle
M = mol/V = 0.00390030977/(54.8*10^-3) = 0.071173 M
(d) If 27.5 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
1 mol of base = 2 mol of acid
mol o fHCl = MV = 27.5*0.125 = 3.4375 mmol of HCl
then
3.4375/2 = 1.71875 mmol of base
mass = mol*MW = (1.71875 *10^-3)(74.093)= 0.127347 g of Ca(OH)2
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