Question

Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 8.0×10⁻²MNaOH.

Chlorous Acid

Benzoic Acid

Answer 1

Solution :-

Part a ) Chlorous acid

Lets assume we have 100 ml of 0.100 M chlorous acid

And we have 8.0E-2 M NaOH

So lets write the balanced reaction equation

HClO2 + NaOH ----- > NaClO2 + H2O

Now lets calculate the volume of the NaOH needed for the reaction

Volume of NaOH needed to reach equivalence point

Volume of NaOH = molarity of acid * volume of acid / molarity of NaOH

                                = 0.100 M * 100 ml / 8.0E-2 M

                               = 125 ml

Now lets calculate the moles of acid

Moles = molarity * volume in liter

Moles of HClO2 =0.100 mol per L * 0.100 L =0.01 mol HClO2

At the equivalence point moles of acid and moles of NaOH are same

Therefore all the acid will get conveted into the conjugate base ClO2-

Therefore now lets calculate new molarity of the ClO2- at total volume

Total volume =100 ml +125 ml = 225 ml = 0.225 L

New molarity of the ClO2- = 0.01 mol / 0.225 L =0.0444 M

Now lets calculate the [OH-] using the pkb of the ClO2-

ClO2- + H2O   ----- > HClO2   + OH-

0.0444 M                   0                    0

-x                                 +x                   +

0.0444-x                     x                     x

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

Kb of the ClO2- = 9.1E-13

Lets put the value of the Kb in the formula

9.1E-13 = [x][x]/[0.0444-x]

Since Kb is very small therefore we can neglect the x from the denominator then we get

9.1E-3 =[x][x]/0.0444

9.1E-3*0.0444 =x²

4.04E-14 = x²

By taking square root of both sides we get

2.01E-7 = x=[OH-]

Now lets calculate the pOH

pOH= -log [OH-]

pOH =-log [2.01E-7]

pOH = 6.69

now lets calculate the pH

pH + pOH = 14

therefore pH= 14 – pOH

pH = 14 – 6.69

pH= 7.31

part 2 ) benzoic acid

lets assume we have 100 ml of 0.100 M benzoic acid solution

mole ratio of the benzoic acid to NaOH is 1 : 1

PhCOOH + NaOH ----- > PhCOONa + H2O

Therefore the volume of the NaOH needed as we calculated in the part a because mole ratio is same and concentrations are also same

There fore volume of NaOH needed = 125 ml

After the reaction all the benzoic acid will convert into the benzoate ions which will react with water to form OH- ions therefore the concentration of the benzoate ion after the reaction is 0.0444 M

PhCOO- + H2O ---- > PhCOOH + OH-

0.0444 M                   0                    0

-x                                 +x                   +

0.0444-x                     x                     x

Kb = [PhCOOH][OH-]/[PhCO2-]

Kb of the PhCOO- = 1.59E-10

Lets put the values in the formula

1.59E-10 = [x][x]/[0.0444-x]

Since Kb is very small therefore we can neglect the x from the denominator then we get

1.59E-10 =[x][x]/0.0444

1.59E-10*0.0444 =x²

7.01E-12 = x²

By taking square root of both sides we get

2.65E-6 M= x=[OH-]

Now lets calculate the pOH

pOH= -log [OH-]

pOH =-log [2.65E-6]

pOH = 5.58

now lets calculate the pH

pH + pOH = 14

therefore pH= 14 – pOH

pH = 14 – 5.58

pH= 8.42