Question

(a) How many milliliters of 0.155 M HCl are needed to neutralize completely 45.0 mL of 0.101 M Ba(OH)₂ solution?

(b) How many milliliters of 3.50 M H₂SO₄ are needed to neutralize 75.0 g of NaOH?

(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?

(d) If 27.5 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?

Answer 1

The reaction between Ba(OH)2 and HCl is

Ba(OH)2 + 2HCl-------> BaCl2 + 2H2O

1 mole of Ba(OH)2 requires 2 mole of HCl for neutralization.

Moles of Ba(OH)2 in 45ml of 0.101M= Molarity* Volume (L) 0.101*45/1000=0.004545 moles

Hence moles of HCl required= 2*0.004545=0.00909

Molarity of HCl given =0.155M

Volume of HCl required= moles/Molarity= 0.00909/0.155=0.058645 L =0.058645*1000ml = 58.64 ml

2.

How many milliliters of 3.50 M H₂SO₄ are needed to neutralize 75.0 g of NaOH?

2NaOH + H2SO4-------> Na2SO4 + 2H2O

moles of NaOH in 75 gm = Mass/Molar mass = 75/40 =1.875 moles

2Moles of NaOH reacts with 1 mole of H2SO4

1.875 moles of NaOH requires 1.875/2= 0.9375 moles of H2SO4

molarity of H2SO4= 3.5M, Volume of H2SO4= moles/Molarity =0.9375/3.5= 0.267L= 0.267*1000=267 ml

3.

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?

BaCl2 + Na2SO4------>BaSO4 +2NaCl

mass of Na2SO4= 544mg= 544/1000 = 0.544 gm

moles of Na2SO4= mass/Molar mass = 0.544/142= 0.003831

molarity = moles/Volume (L)= 0.003831*1000/56.8=0.067M

4. Ca(OH)2 + 2HCl ------> CaCl2

moles of HCl = 0.375*27.5/1000 =0.010 ,moles of Ca(OH)2 required= 0.010/2= 0.0052

mass of Ca(OH)2= moles* molar mass =0.0052*74=0.3848 gm