Below is a pi I curve for the titration of a weak acid(HA) with a strong base (N

Question

Below is a pi I curve for the titration of a weak acid(HA) with a strong base (NaOH). Write out the general titration reaction. List the major species in solution for each of the following volumes,, indicate how you would go about calculating the pH of the solution at these volumes. On the figure itself, indicate the half way point and the equivalence point. The volumes are 0, 10,20, 40 and 50mL of titrant. 0 mL: At 0 mL no base has been added, the only ions in solution is from the dissociation of the weak acid HA to give H+ and A-. To calculate the pH at this point you would solve it like a weak acid problem

Explanation / Answer

Titration curve

a) General titration reaction,

HA + NaOH ---> NaA + H2O

b) pKa of the weak acid from the curve is pH correesponding to volume 20 ml of NaOH added.

pKa = 4.75

This is the half equivalence point when pH = 4.75

Equivalence point is at Volume of base NaOH = 40 ml, pH around = 8.2

pH calculation when,

i) 0 ml NaOH added

Ka = [H+][A-]/[HA]

Calculate [H+] and then pH

Major species present = HA

ii) 10 ml NaOH

pH = pKa + log(base/acid)

some part of weak acid is neutralized and forms NaA which is taken as the base in the above equation.

Major species present = HA and minor NaA

iii) 20 ml NaOH added

Half equivalence point

pH = pKa

major species present : [HA] = [NaA]

iv) 40 ml NaOH

Equivalence point

All of weak acid is neutralised

Kb = [HA][OH-]/[A-]

calculate [OH-] an then [H+] for pH

major species present = A-

v) 50 ml NaOH

pH = 14 - pOH

[OH-] = excess NaOH = 10 ml

major species present = OH-               

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