Question

Calculate the pH at the equivalence point in titrating 0.110 M solutions of each of the following acids with a solution 0.090 M in NaOH.

HClO2

HBr

Answer 1

Part a

In HClO2

Basis - 10.0 mL of 0.110 M HClO2

moles of HClO2 = Molarity x volume

= (0.110 mol/L)(10.0 mL x 1L/1000 mL ) = 0.0011 moles

The balanced reaction

HClO2 + NaOH ==> NaClO2 + H2O.

Molar ratio HClO2 : NaOH = 1 : 1

Moles of NaOH = 0.0011 moles

Volume of NaOH = 0.0011 moles / 0.09M

= 0.0122 L

Total volume = 0.0122 + 0.01 = 0.0222 L

Molarity of NaClO2 = moles NaClO2 /volume

= 0.0011 / 0.0222

= 0.0495 M

at the equivalence point

ICE TABLE

Molarity . . . . .ClO2- + H2O <==> HClO2 + OH-
Initial . . . . . . 0.0495   
Change . . . . . . .-x . . . . . . . . . . . . . . . .x . . . . .x
Equilibrium . .0.0495-x . . . . . . . . . . . . . x . . . . .x

Kb of ClO2- = Kw / Ka

= (1.0 * 10^-14) / (1.1 * 10^-2)

= 9.1 * 10^-13

Kb = [HClO2][OH-] / [ClO2-]

9.1 * 10^-13 = (x)(x) / (0.0495-x)

x << 0.0495

x^2 / 0.0495 = 9.1 * 10^-13
x^2 = 4.51 * 10^-14
x = 2.12*10^-7 = [OH-]
pOH = -log [OH-] = -log (2.12 * 10^-7) = 6.67
pH = 14.00 - pOH = 14.00 - 6.67 = 7.33

Part b

HBr is a strong acid

NaOH is a strong base.

At equivalence point, due to neutralization effect

pH =7