Question

Be sure to answer all parts. A 24.0−mL solution of 0.110 M CH3COOH is titrated with a 0.220 M KOH solution. Calculate the pH after the following additions of the KOH solution: (a) 10.0 mL (b) 12.0 mL (c) 15.0 mL

Answer 1

Solution :-

Addition of the KOH to the acetic acid solution forms the buffer

24.0 ml of 0.110 M CH3COOH

KOH molarity = 0.220 M

CH3COOH + OH^-   ---- > CH3COO^- + H2O

Part a) Calculating the pH after adding 10 ml KOH

Lets first calculate the moles of acid and base

Mole =molarity x volume in liter

Moles of CH3COOH = 0.110 mol per L * 0.0240 L = 0.00264 mol

Moles of KOH = 0.220 mol per L * 0.010 L = 0.00220 mol

Moles of KOH are less than moles of CH3COOH

Therefore after the reaction moles of CH3COOH remain are calculated as

Moles of CH3COOH after reaction = 0.00264 mol – 0.00220 mol = 0.00044 mol

Moles of conjugate formed = 0.0022 mol

Total volume = 24.0 ml + 10.0 ml = 34.0 ml = 0.0340 L

Lets calculate the new molarity of the acid and conjugate base

Molarity = moles /volume in liter

New molarity of [CH3COOH]= 0.00044 mol / 0.0340 L = 0.01294 M

New molarity of [CH3COO-] = 0.0022 mol / 0.0340 L = 0.0647 M

Now lets calculate the pH using the Henderson equation

pH= pka + log [CH3COO-]/[CH3COOH]

pka of acetic acid is 4.74

pH= 4.74 + log [0.0647]/[0.01294]

pH= 5.44

Therefore pH= 5.44 after adding 10.0 ml KOH

part b)after adding 12 ml KOH

Moles of CH3COOH = 0.110 mol per L * 0.0240 L = 0.00264 mol

Moles of KOH = 0.220 mol per L * 0.012 L = 0.00264 mol

Here moles of acid and moles of base are same therefore the reaction is reached to equivalence point therefore all the acid is converted to conjugate base CH3COO- which react with water to form the OH^-

New molarity of the conjugate base is calculate at total volume (24.0 ml + 12.0 ml = 36.0 ml = 0.0360 L)

[CH3COO-] = 0.00264 mol / 0.0360 L = 0.0733 M

Using the Kb of the CH3COO- we can find the equilibrium concentration of the OH^-

CH3COO^- + H2O   ---- > CH3COOH + OH^-

0.0733                                       0                0

-x                                               +x              +x

0.0733-x                                     x                  x

Kb =[CH3COOH][OH-]/[CH3COO-]

5.56*10^-10 = [x][x]/[0.0733]

5.56*10^-10 * 0.0733 = x^2

4.08*10^-11 = x^2

Taking square root on both sides we get

6.38*10^-6 M= x = [ OH-]

Now lets calculate the pOH

pOH= -log [OH-]

pOH= - log [ 6.38*10^-6]

       = 5.20

pH + pOH = 14

pH= 14 – pOH

    = 14 – 5.20

      = 8.80

Therefore the pH is 8.80 after addition of the 12.0 ml KOH

Part c)

After adding 15 ml KOH

Moles of CH3COOH = 0.110 mol per L * 0.0240 L = 0.00264 mol

Moles of KOH = 0.220 mol per L * 0.015 L = 0.00330 mol

Here moles of KOH are more than moles if acid

Therefore after the completion of reaction some KOH is remain in the solution

Excess moles of KOH= 0.0033mol – 0.00264mol = 0.00066 mol

Total volume = 24.0 ml + 15.0 ml = 39.0 ml = 0.0390 L

New molarity of the KOH = 0.00066 mol / 0.03690 L

                                          = 0.0169 M

KOH is strong base therefore the molarity of the KOH is same as molarity of the OH^-

Therefore [OH^- ] = 0.0169 M

Now lets calculate the pOH

pOH= -log [OH-]

        = -log [0.0169]

        = 1.77

pH= 14 – pOH

    = 14 – 1.77

    = 12.23

Therefore the pH= 12.23 after adding 15.0 ml KOH