Question

3. (A)

A 25.0 mL sample of 0.103 M Ba(OH)₂ was titrated with 0.188 M HClO₄. Calculate the pH of the solution before the addition of any titrant and after the addition of 15.6, 27.4, and 38.1 mL of titrant.

Answer 1

first,the eqution of neutralization is given

Ba(OH)2 + 2HCLO4 --> Ba(ClO4)2 + 2H2O

now...

initially, we have

mmol of Ba(OH)2 = MV = 25*0.103 = 2.575 mmol of Ba(OH)2

mmol of OH- = 2*Ba(OH)2 = 2*2.575 = 5.15 mmol of OH-

a)

mmol of HClO4 added = Macid*Vacid = 0.188*15.6 = 2.9328 mmol

so..

mmol of OH- left = 5.15 -2.9328 = 2.2172 mmol o fOH-

Vtotal = V1+V2 = 25+15.6 = 40.6 mL

[OH-] = mmol/V = 2.2172 /40.6 = 0.054610 M

pOH = -log(0.054610) = 1.2627

pH = 14-1.2627 = 12.7373

b)

mmol of HClO4 added = Macid*Vacid = 0.188*27.4= 5.1512 mmol

so..

mmol of H+ left = 5.15 -5.1512= 0.0012 mmol of H+

Vtotal = V1+V2 = 25+27.4= 52.4mL

[H+] = mmol/V = 0.0012 /52.4= 0.0000229 M

pH = -log(0.0000229 = 4.640

c)

mmol of HClO4 added = Macid*Vacid = 0.188*38.1= 7.1628mmol

so..

mmol of H+ left = 7.1628-5.15= 2.0128 mmol of H+

Vtotal = V1+V2 = 25+38.1= 63.1mL

[H+] = mmol/V = 2.0128/63.1= 0.03189 M

pH = -log(0.03189= 1.4963