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

Answer 1

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