(a) How many milliliters of 0.145 M HCl are needed to neutralize completely 35.0

Question

(a) How many milliliters of 0.145 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 25.0 g of NaOH? ____mL (c) If 54.8 mL of BaCl2 solution is needed to precipitate all the sulfate in a 534 mg sample of Na2SO4 (forming BaSO4), what is the molarity of the solution? _______M (d) If 37.5 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

Explanation / Answer

(a) How many milliliters of 0.145 M HCl are needed to neutralize completely 35.0 mL of 0.101 M Ba(OH)2 solution?

     millimoles of HCl = molarity x volume (ml)

                                  = 0.145 x V

millimoles of Ba(OH)2 = 0.101 x 35

                                     = 3.535

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

2 x HCl millimoles = Ba(OH)2 millimoles

2 x 0.145 x V = 3.535

V =12.2 ml

HCl volume = 12.2 ml

(b) How many milliliters of 3.50 M H2SO4 are needed to neutralize 25.0 g of NaOH? ____mL

   NaOH moles = mass / molar mass

                         = 25 / 40

                         = 0.625

moles of H2SO4 = 3.5 x V

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

1 mol H2So4 = 2 mol NaOH

3.5 x V = 2 x 0.625

V = 0.357 L

V = 357 ml

volume of H2SO4 = 357 ml

(c) If 54.8 mL of BaCl2 solution is needed to precipitate all the sulfate in a 534 mg sample of Na2SO4 (forming BaSO4), what is the molarity of the solution? _______M

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

Na2SO4 moles = 534 x 10^-3 / 142

                         = 3.76 x 10^-3

solution molarity = moles / volume (L)

                           = 3.76 x 10^-3   x 1000 / 54.8

                           = 0.069 M

d) If 37.5 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

moles of HCl = 37.5 x 0.250 / 1000

                       = 9.38 x 10^-3

2HCl + Ca(OH)2 ---------------------> CaCl2 + 2HO

Ca(OH)2 moles = 1/2 HCl moles

                          = 1/2 x 9.38 x 10^-3

                          = 4.69 x 10^-3

Ca(OH)2 molar mass = 74 g/mol

mass of Ca(OH)2 = moles x molar mass

                             = 4.69 x 10^-3   x 74

                            = 0.347 g

weight of Ca(OH)2 = 0.347 g

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