Question 19.1: Calculating the Effect of Added H3O^+ or OH^− on Buffer pH P......

Plan For each part, we know, or can find, [HClO]_{\text{init}} and [ClO^−]_{\text{init}}. We know the K_a of HClO  (2.9×10^{−8}) and need to find [H_3O^+] at equilibrium and convert it to pH. (a) We use the given concentrations of buffer components (0.50 M HClO and 0.60 M ClO^−) as the initial values. As in earlier problems, we assume that x, the [HClO] that dissociates, which equals [H_3O^+], is so small relative to [HClO]_{\text{init}} that it can be neglected. We set up a reaction table, solve for x, and check the assumption. (b) and (c) We assume that the added OH^− or H_3O^+ reacts completely with the buffer components to yield new [HClO]_{\text{init}} and [ClO^−]_{\text{init}}, and then the acid dissociates to an unknown extent. We set up two reaction tables. The first summarizes the stoichiometry of adding strong base or acid. The second summarizes the dissociation of the new [HClO]_{\text{init}}, and we proceed as in part (a) to find the new [H_3O^+]. Since there is a volume change due to the added HCl in part (c), we must calculate the new initial concentrations of the buffer components.

Solution (a) The original pH: [H_3O^+] in the original buffer. Setting up a reaction table (Table 1) with x  =  [HClO]_{\text{dissoc}}  =  [H_3O^+] (as in Chapter 18, we assume that [H_3O^+] from H_2O is negligible and disregard it):

Making the assumption and finding the equilibrium [HClO] and [ClO^−]: With K_a small, x is small, so we assume
            [HClO]  =  0.50  M  −  x  ≈  0.50  M    and     [ClO^−]  =  0.60  M  +  x  ≈  0.60  M
Solving for x ([H_3O^+] at equilibrium):

Checking the assumption:
           \frac{2.4×10^{−8}  M}{0.50  M}  ×  100  =  4.8×10^{−6}\%  <  5\%
The assumption is justified, and we will use the same assumption in parts (b) and (c). Also, according to the other criterion (see Sample Problem 18.8),

Calculating pH:
          pH  =  −\log  [H_3O^+]  =  −\log  (2.4×10^{−8})  =  7.62
(b) The pH after adding base (0.020 mol of NaOH to 1.0 L of buffer). Finding [OH^{−}]_{\text{added}}:

Setting up a reaction table (Table 2) for the stoichiometry of adding OH^− to HClO (notice that the reaction between weak acid and strong base goes to completion):

Setting up a reaction table (Table 3) for the acid dissociation, using these new initial concentrations. As in part (a), x  =  [HClO]_{\text{dissoc}}  =  [H_3O^+]:

Making the assumption that x is small, and solving for x:
          [HClO]  =  0.48   M  −  x  ≈  0.48 M     and     [ClO^−]  =  0.62  M  +  x  ≈  0.62  M
          x  =  [H_3O^+]  =  K_a  ×  \frac{[HClO]}{[ClO^−]}  ≈  (2.9×10^{−8})  ×  \frac{0.48}{0.62}  =  2.2×10^{−8}  M

Calculating the pH:
          pH  =  −\log  [H_3O^+]  =  −\log  (2.2×10^{−8} )  =  7.66
The addition of strong base increased the concentration of the basic buffer component at the expense of the acidic buffer component. Note especially that the pH increased only slightly, from 7.62 to 7.66.

(c) The pH after adding acid (30.0 mL of 1.5 M HCl to 0.50 L of buffer). Use Equation 4.2 to find the new [HClO]_{\text{init}} and [ClO^−]_{\text{init}} and [H_3O^+]_{\text{added}} after combining the HCl and buffer solutions; the new volume is 0.50 L + 0.0300 L = 0.53 L:

         M_{\text{dil}}  ×  V_{\text{dil}}  =  \text{amount (mol) of solute }=  M_{\text{conc}}  ×  V_{\text{conc}}      (4.2)

         [HClO]_{\text{init}}  =  \frac{M_{\text{conc}}  ×  V_{\text{conc}}}{V_{\text{dil}}}  =  \frac{0.50  M  ×  0.50  L}{0.53  L}  =  0.47  M  HClO
         [ClO^−]_{\text{init}}  =  \frac{M_{\text{conc}}  ×  V_{\text{conc}}}{V_{\text{dil}}}  =  \frac{0.60  M  ×  0.50  L}{0.53  L}  =  0.57  M  ClO^−
         [H_3O^+]_{\text{added}}  =  \frac{M_{\text{conc}}  ×  V_{\text{conc}}}{V_{\text{dil}}}  =  \frac{1.5  M  ×  0.0300  L}{0.53  L}  =  0.085  M  H_3O^+

Now we proceed as in part (b), by first setting up a reaction table (Table 4) for the stoichiometry of adding H_3O^+ to ClO^− (notice that the reaction between strong acid and weak base goes to completion):

The reaction table (Table 5) for the acid dissociation, with x  =  [HClO]_{\text{dissoc}}  =  [H_3O^+] is

Making the assumption that x is small, and solving for x:
         [HClO]  =  0.56  M  −  x  ≈  0.56  M and [ClO^−]  =  0.48  M  +  x  ≈  0.48  M
         x  =  [H_3O^+]  =  K_a  ×  \frac{[HClO] }{[ClO^−]}  ≈  (2.9×10^{−8})  ×  \frac{0.56}{0.48}  =  3.4×10^{−8}  M
Calculating the pH:

The addition of strong acid increased the concentration of the acidic buffer component at the expense of the basic buffer component and lowered the pH only slightly, from 7.62 to 7.47.
Check The changes in [HClO] and [ClO^−] occur in opposite directions in parts (b) and (c), which makes sense. The pH increased when base was added in (b) and decreased when acid was added in (c).

Comment In part (a), we justified our assumption that x can be neglected. Therefore, in parts (b) and (c), we could have used the “Final” values from the last line of the stoichiometry reaction tables directly for the ratio of buffer components; that would have allowed us to dispense with a reaction table for the acid dissociation. In subsequent problems in this chapter, we will follow this more straightforward approach.