O-0.1 points 0/4 Submissions Used My Notes Ask Your Teacher (a) How many millili
ID: 1038968 • Letter: O
Question
O-0.1 points 0/4 Submissions Used My Notes Ask Your Teacher (a) How many milliliters of 0.155 M HCI are needed to neutralize completely 25.0 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 55.8 mL of BaCl2 solution is needed to precipitate all the sulfate in a 534 mg sample of Na2S04 (forming BaSO4), what is the molarity of the solution? (d) If 37.5 mL of 0.375 M HCI solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solution?Explanation / Answer
1) 2 HCl + Ba(OH)2 ---------------- BaCl2 + 2 H2O
2 mole 1 mole
HCl Ba(OH)2
M1= 0.155M M2= 0.101M
V1= V2= 25.0ml
for neutralisation reactions
M1V1/n1= M2V2/n2
0.155xV1/2 = 0.101x25.0/1
V1=32.58 ml
volume of HCl= 32.58 ml
2) mass of NaOH= 75.0grams
molar mass of NaOH= 40.0grams
number of moles of NaOH= 75.0/40.0= 1.875 moles
H2SO4 + 2 NaOH---------- Na2SO4 + 2 H2O
1 mole 2 mole
according to equation
2 mole of NaOH= 1 mole H2SO4
1.875 mole of NaOH= ?
= 1.875x1/2= 0.9375 moles
number of moles of H2SO4= 0.9375 moles
M= 1.50M
Molarity = number of moles/volume in L
volume= 0.9375/1.50=0.625L
Volume = 0625L
Volume = 625 ml
d)
l
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