Question

A certain weak acid, HA, with a Ka value of 5.61�10?6, is titrated with NaOH.

A) A solution is made by titrating 7.00mmol (millimoles) of HA and 2.00mmol of the strong base. What is the resulting pH?

B) More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 38.0mL ?

WILL RATE IF BOTH ANSWERS ARE CORRECT WITH STEPS SHOWN

Answer 1

a) on addition of base the acid will react with it to form salt

HA + NaOH ---> NaA + H2O

Initial moles of acid present = 7mmol

Moles of base added =2 mmol

Moles of acid reacted = 2 millimoles

Moles of salt formed = 2mmoles

Moles of acid left = 7-2 = 5 millmoles

It will form a buffer for which we can calculate the pH using Hendersen Hassalbalch equation

pH = pKa + log [salt] / [acid]

pKa = -log Ka = 5.25

pH = 5.25 + log [2] / [5] = 4.85

b) at equivalence the millimoles of salt formed = 7 millimoles

[Salt] = [A-] = Millimoles / Volume in mL = 7 / 38 = 0.184 M

It will hydrolyze as

                   A- + H2O -- --> AH + OH-

Initial           0.184                0        0

Change        -x                    +x      +x

Equilibrium   0.184-x              x       x

Kb = Kw / Ka = [HA] [ OH-] / [A-] = x^2 / (0.184-x) = 10^-14 / 5.61 X 10^-6

we may ignore x in denominator

0.178 X 10^-8 = x^2 / (0.184)

x = 1.809 X 10^-5 = [OH-]

pOH = -log[OH-] = 4.74

pH = 14 - pOH = 9.26