Question

A 120.0-mL aliquot of 0.114 M weak base B (pKb = 4.58) was titrated with 1.14 M HClO4. Find the pH at the following volumes of acid added: Va = 0.00, 1.30, 6.00, 11.00, 11.90, 12.00, 12.10, and 17.00 mL. (Assume Kw = 1.01 ✕ 10−14.)

Answer 1

Solution :-

Calculation of pH at 0.00 ml addition of HClO4

Initially the base is present therefore using the concentration of the base we can find the OH^- concentration

Pkb= 4.58

Kb= antilog[-pkb]

    = antilog[-4.58]

    = 2.63*10^-5

   B +   H2O   ----- > BH^+   + OH^-

0.114                          0               0

-x                              +x                +x

0.114-x                      x                  x

Kb=[BH+][OH-]/[B]

2.63*10^-5 = [x][x]/[0.114-x]

Since the kb is small therefore we can neglect the x from denominator

2.63*10^-5 = [x][x]/[0.114]

2.63*10^-5 * 0.114 = x^2

3.0*10^-6 = x^2

Taking square root on both sides we get

1.73*10^-3 = x

pOH= -log[OH^-]

pOH= -log[1.73*10^-3]

pOH= 2.76

pH = 14 – pOH

pH= 14 – 2.76

pH= 11.24

calculation of pH at 1.30 ml addition of HClO4

Addition of the strong acid to the weak base solution forms the buffer

Initial moles of base = molarity x volume in liter

                                     = 0.114 mol per L * 0.120 L

                                    = 0.01368 mol

Initial moles of acid = 1.14 mol per L * 0.00130 L = 0.001482 mol

After the reaction moles of base remain = 0.01368 mol – 0.001482 mol = 0.01220 mol

Moles of conjugate acid formed = 0.001482 mol

Total volume = 120 ml + 1.30 ml = 121.3 ml = 0.1213 L

Using the ratio of the conjugate acid and base we can find the pH using the Henderson equation

pH= pka+ log[base]/[acid]

pka= 14- pkb

       = 14 – 4.58

        = 9.42

pH= 9.42 + log[0.01220]/[0.001482]

pH= 10.33

Calculation of pH at the addition of 6.00 ml HClO4

Initial moles of base = 0.114 mol per L * 0.120 L = 0.01368 mol

Initial moles of acid = 1.14 mol per L * 0.0060 L = 0.00684 mol

Moles of base remain after reaction = 0.01368 mol – 0.00684 mol = 0.00684 mol

Moles of conjugate acid formed = 0.00684 mol

Here moles of conjugate acid and base are same therefore reaction is reached at the half equivalence point

At the half equivalence point pH= pka

Therefore

pH= pka

pH= 9.42

calculation of pH at the addition of 11.00 ml HClO4

Initial moles of base = 0.114 mol per L * 0.120 L = 0.01368 mol

Initial moles of acid = 1.14 mol per L * 0.0110 L = 0.01254 mol

Moles of base remain after reaction = 0.01368 mol – 0.01254 mol = 0.00114 mol

Moles of conjugate acid formed = 0.01254 mol

pH= pka+ log [base]/[acid]

pH= 9.42 + log[0.00114]/[0.01254]

pH= 8.38

calculation of pH at the 11.90 ml addition of HClO4

Initial moles of base = 0.114 mol per L * 0.120 L = 0.01368 mol

Initial moles of acid = 1.14 mol per L * 0.0119 L = 0.01357 mol

Moles of base remain after reaction = 0.01368 mol – 0.01357 mol = 0.00011 mol

Moles of conjugate acid formed = 0.01357 mol

pH= pka+ log [base]/[acid]

pH= 9.42 + log[0.0011]/[0.013577]

pH= 7.33

calculation of the pH at 12.00 ml addition of HClO4

Initial moles of base = 0.114 mol per L * 0.120 L = 0.01368 mol

Initial moles of acid = 1.14 mol per L * 0.0120 L = 0.01368 mol

Here moles of acid and moles of base are same therefore all the base is converted to conjugate acid

Total volume = 120 ml + 12.0 ml = 132.0 ml = 0.132 L

New molarity of conjugate acid = moles/ volume

                                                        = 0.01368 mol / 0.132 L

                                                       = 0.1036 M

BH^+   +H2O ---- > B   + H3O^+

0.1036                        0        0

-x                                 +x    +x

0.1036-x                     x       x

Ka=kw/kb

    =1*10^-14 / 2.63*10^-5

    = 3.80*10^-10

Ka=[B][H3O+]/[BH+]

3.80*10^-10 = [x][x]/[0.1036-x]

Since the ka is very small therefore we can neglect the x from denominator

3.80*10^-10 = [x][x]/[0.1036]

3.80*10^-10 * 0.1036 = x^2

3.94*10^-11 = x^2

Taking square root on both sides we get

6.27*10^-6 = x =[H3O+]

pH= -log[H3O+]

pH= -log[6.27*10^-6]

pH= 5.20

calculation of pH at the 12.10 ml addition of HClO4

Initial moles of base = 0.114 mol per L * 0.120 L = 0.01368 mol

Initial moles of acid = 1.14 mol per L * 0.0121 L = 0.01379 mol

Here moles of acid are more than moles of base

Therefore moles of acid remain after reaction = 0.01379 mol – 0.01368 mol = 0.00011 mol

Total volume = 120 ml + 12.10 ml = 132.10 ml = 0.1321 L

New molarity of acid = 0.00011 mol / 0.1321 L = 0.000833 M

HClO4 is strong acid therefore we can use the direct acid concentration to calculate the pH

pH= -log[H^+]

pH= -log[0.000833]

pH= 3.08

calculation of pH at 17.00 ml addition of HClO4

Initial moles of base = 0.114 mol per L * 0.120 L = 0.01368 mol

Initial moles of acid = 1.14 mol per L * 0.0170 L = 0.01938 mol

Moles of acid remain after reaction = 0.01938 mol – 0.01368 mol = 0.0057 mol

Total volume = 120 ml + 17.0 ml = 137.0 ml = 0.137 L

New molarity of acid = 0.0057 mol / 0.137 L = 0.0416 M

pH= -log[H^+]

pH= -log[0.0416]

pH= 1.38