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 following volumes of HCl have been added. (a) 0.0 mL WebAssign will check your answer for the correct number of significant figures. 13.447 Correct: Your answer is correct. (b) 15.0 mL WebAssign will check your answer for the correct number of significant figures. 12.25 Incorrect: Your answer is incorrect.

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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