(a) How many milliliters of 0.165 M HCl are needed to neutralize completely 35.0

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 2.50 M H2SO4 are needed to neutralize 75.0 g of NaOH?
mL


(c) If 56.8 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 0.250 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

Please show steps with units of measurement

Explanation / Answer

Moles Ba(OH)2 = 0.101 M x 0.0350 = 0.003535
Moles OH- = 2 x 0.003535 =0.007070
= moles HCl needed
V = 0.007070 / 0.165 M = 0.0428 L=> 42.8 mL

75.0 g NaOH / 40.0 g/mol = 1.875 mol
2NaOH + H2SO4 --> Na2SO4 + 2H2O
1 mol H2SO4 will neutralize 2 mol NaOH
You require 1.875 / 2 = 0.9375 mol H2SO4
liters x molarity = moles
? liters x 2.50 M = 0.9375 mol
? liters = 0.375 L = 375 mL H2SO4

0.544 g Na2SO4 @142.04 g/mol = 0.003830 moles Na2SO4

0.003830 moles Na2SO4 reacts with 0.003830 moles BaCL2

0.003830 mol BaCl2 / 0.0558 litres = 0.06864 Molar

your answer: 0.0686 Molar BaCl2

0.0475 L of 0.25 mol/litre HCl = 0.011875 moles HCl

0.011875 moles HCl reacts @ 1 mol Ca(OH)2 / 2 mol HCl = 0.0059375 mol Ca(OH)2

find grams using molar mass:
0.0059375mol Ca(OH)2 @ 74.09 g/mol = 0.5399 grams Ca(OH)2

your answer is: 0.4399 grams Ca(OH)2

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.