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Serga [27]
3 years ago
10

PLEASE HELP ME.!

Chemistry
2 answers:
katen-ka-za [31]3 years ago
6 0
The answer is Anaphase II
Karo-lina-s [1.5K]3 years ago
4 0

Answer:

Anaphase II

Explanation:

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3 years ago
Consider the titration of 82.0 mL of 0.140 M Ba(OH)2 by 0.560 M HCl. Calculate the pH of the resulting solution after the follow
oksano4ka [1.4K]

Answer:

a) pH = 13.447

b) pH = 13.176

Explanation:

Ba (OH)₂ is a strong base and ionizes completely in solution to give barium amend hydroxide ions.

The equation of the dissociation of Ba(OH)₂ is given below:

Ba(OH)₂ ----> Ba2+ + 2OH-

1 mole of Ba(OH)₂ produces 2 Moles of OH- ions

a) Before the addition of HCl, i.e.,when 0.00 mL of HCl has been added;

Concentration of hydroxide ions, [OH-] = 0.140 x 2 = 0.a) [OH-] = 0.100 x 2 = 0.280

pOH = -log [OH-]

pOH = -log (0.280)

pOH = 0.553

pH = 14 - 0.553= 13.447

b) pH when 15.0 mL HCl has been added

Moles Ba(OH)₂ = concentration × volume

Volume of Ba(OH)₂ = 82.0 mL = 0.082 L, concentration = 0.140 M

moles of Ba(OH)₂ = 0.082 x 0.140 = 0.01148 moles

moles OH- produced by 0.01148 moles of Ba(OH)₂ = 2 x 0.01148= 0.02296

moles HCl = 0.0150 L x 0.560 = 0.0084

moles of H+ produced by 0.0084 HCl = 0.0084

0.0084 H+ willnreact with 0.0084 moles of OH-

moles OH- left after the reaction = 0.02296 - 0.0084 = 0.01456 moles

total volume of new solution = (82 +15) mL = 97 mL => 0.097 L

Concentration of OH- ions = moles / volume

[OH-] = 0.01456 / 0.097 = 0.1501

pOH = -log [OH-]

pOH = -log (0.150)

pOH = 0.824

pH = 14 - 0.824

pH = 13.176

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