Question

CHM2046L: General Chemistry and Qualitative Analysis II; Fall 2019 Lab # 9: EQUILIBRIUM: Calculating pH and Buffer Capacity Post-lab # 9: ( Due Monday, November 7t, 2019) Last name First Name 1. Define buffer capacity. 2. Determine the Kb for CN at 25°C. The Ka for HCN is 4.9 x 10-10, 3. Determine the [H30] in a 0.265 M HCIO solution. The Ka of HCIO is 2.9 x 10-8 rco 4. Determine the pH of a 0.461 M C6H5CO2H M solution if the Ka of C6H5CO2H is 6.5 x 10-5 bl oa Ioptuios 5. Calculate the hydronium ion concentration in an aqueous solution with a pH of 9.85 at 25°C.

Answer 1

1) The property of a buffer solution to resist alteration in its pH value is known as buffer capacity. It has been found that if the ratio [Salt]/[Acid]   or [Salt]/[Base] is unity, the pH of a particular buffer does not change at all.

Buffer capacity is defined quantitatively as “number of moles of acid or base added in one litre of solution so as to change the pH by unity, i.e.
(∅)=(No.of moles of acid or base added to litre)/(change in pH)

2)

Ka*Kb = 1*10^-14

    Ka = 4.9*10^-10

    Kb = ?

(4.9*10^-10)*Kb = 1*10^-14

Kb = 2.04*10^-5

3)

            HClO(aq) + H2O(l) <-----> H3O+(aq) + ClO-(aq)

Initial     0.265 M                      -         -

change        -x                         +x         +x

equil      0.265-x                        x         x

Ka = [H+][ClO-]/[HClO]

2.9*10^-8 = (x*x)/(0.265-x)

x = 8.76*10^-5

at equilibrium,

[H3O+] = x = 8.76*10^-5 M

4)

          C6H5COOH(aq) + H2O(l) <-----> H3O+(aq) + C6H5COO-(aq)

Initial       0.461 M                          -            -

change         -x                             +x            +x

equil        0.461-x                           x             x

Ka = [H+][ClO-]/[HClO]

6.5*10^-5 = (x*x)/(0.461-x)

x = 5.44*10^-3

at equilibrium,

[H3O+] = x = 5.44*10^-3 M

pH = -log[H3O+]

   = -log(5.44*10^-3)

   = 2.26

5) pH = -log[H+]

9.85 = -log[H+]

[H+]= 1.41*10^-10 M