12. -10.1 points 0/4 Submissions Used (a) How many milliliters of 0.155 M HCI ar

Question

12. -10.1 points 0/4 Submissions Used (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? (b) How many milliliters of 2.50 M H2S04 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 534 mg sample of Na2SO4 (forming BaSO4), what is the molarity of the solution? (d) If 47.5 mL of 0.250 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

a) The reaction is given as :

2HCl + Ba(OH)2 = BaCl2 + 2H2O

The volume of HCl can be calculated by the fomula ,

M1V1/n1 = M2V2/n2

0.155 x V / 2 = 0.101 x 25 / 1

V = 32.58 mL

So the volume of HCl required will be 32.58 mL

b) The reaction is given as :

H2SO4 + 2NaOH = Na2SO4 + 2H2O

Number of moles of NaOH = 75.0 / 40 = 1.875 moles

So the number of moles of H2SO4 = 0.9375 moles

Volume = 0.9375 / 2.50

= 0.375 L or 375 mL

c) The reaction is given as :

BaCl2 + Na2SO4 = BaSO4 + 2NaCl

Number of moles of Na2SO4 = 0.534 / 142.04 = 0.00376 moles

Molarity of BaCl2 = 0.00376 / 0.0568
= 0.0662 M

d) The reaction is given as :

2HCl + Ca(OH)2 = CaCl2 + 2H2O

Number of moles of HCl = 0.0475 x 0.250 = 0.011875 moles

So the number of moles of Ca(OH)2 = 0.011875 / 2 = 0.0059375 moles

Amount of Ca(OH)2 = 0.0059375 x 74.093 = 0.44 grams

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