Calculating the pH of a Buffer When a Strong Acid or Strong Base Is Added Calculate the pH of 75 mL of the buffer solution described in Example 17.10 (0.10 M HC2H3O2 and 0.20 M NaC2H3O2) to which 9.5 mL of 0.10 M hydrochloric acid is added. Compare the pH change with what would occur if this amount

Question

Calculating the pH of a Buffer When a Strong Acid or Strong Base Is Added

Calculate the [latex]\mathrm{pH}[/latex] of [latex]75 \mathrm{~mL}[/latex] of the buffer solution described in Example 17.10 (0.10 [latex]M \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}[/latex] and [latex]0.20 M \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}[/latex] ) to which [latex]9.5 \mathrm{~mL}[/latex] of [latex]0.10 M[/latex] hydrochloric acid is added. Compare the [latex]\mathrm{pH}[/latex] change with what would occur if this amount of acid were added to pure water.

PROBLEM STRATEGY

Do the problem in two parts. First, you assume that the [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex]ion from the strong acid and the conjugate base from the buffer react completely. This is a stoichiometric calculation. Actually, the [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex]ion and the base from the buffer reach equilibrium just before complete reaction. So you now solve the equilib- rium problem using concentrations from the stoichiometric calculation. Because these concentrations are not far from equilibrium, you can use the usual simplifying assumption about [latex]x[/latex].

Accepted Answer

When hydronium ion (from hydrochloric acid) is added to the buffer, it reacts with acetate ion.

[latex]\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}(a q) \longrightarrow \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)[/latex]

Because acetic acid is a weak acid, you can assume as a first approximation that the reaction goes to completion. This part of the problem is simply a stoichiometric calculation. Then you assume that the acetic acid ionizes slightly. This part of the problem involves an acid-ionization equilibrium.

STOICHIOMETRIC CALCULATION You must first calculate the amounts of hydrogen ion, acetate ion, and acetic acid present in the solution before reaction. The molar amount of hydrogen ion will equal the molar amount of hydrochloric acid added, which you obtain by converting the volume of hydrochloric acid, [latex]\mathrm{HCl}[/latex], to moles of [latex]\mathrm{HCl}[/latex]. To do this, you note for [latex]0.10 \mathrm{M} \mathrm{HCl}[/latex] that [latex]1 \mathrm{~L} \mathrm{HCl}[/latex] is equivalent to [latex]0.10 \mathrm{~mol} \mathrm{HCl}[/latex]. Hence, to convert [latex]9.5 \mathrm{~mL}\left(=9.5 imes 10^{-3} \mathrm{~L} ight)[/latex] of hydrochloric acid, you have

[latex] 9.5 imes 10^{-3} \cancel{\mathrm{LHCl} } imes \frac{0.10 \mathrm{~mol } \mathrm{ HCl}}{1 \cancel{\mathrm{LHCl} }}=0.00095 \mathrm{~mol} \mathrm{HCl} [/latex]

Hydrochloric acid is a strong acid, so it exists in solution as the ions [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex]and [latex]\mathrm{Cl}^{-}[/latex]. Therefore, the amount of [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex]added is [latex]0.00095 \mathrm{~mol}[/latex].

The amounts of acetate ion and acetic acid in the [latex]75-\mathrm{mL}[/latex] sample are found in a similar way. The buffer in Example 17.10 contains [latex]0.20 \mathrm{~mol}[/latex] of acetate ion and [latex]0.10 \mathrm{~mol}[/latex] of acetic acid in [latex]1 \mathrm{~L}[/latex] of solution. The amounts in [latex]75 \mathrm{~mL}(=0.075 \mathrm{~L})[/latex] of solution are obtained by converting to moles.

[latex] \begin{aligned} & 0.075 \cancel{ ext { L soln }} imes \frac{0.20 \mathrm{~mol} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-}}{1 \cancel{ ext { L soln }}}=0.015 \mathrm{~mol} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-} \ & 0.075 ext { Lsotn } imes \frac{0.10 \mathrm{~mol} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}}{1 \cancel{ ext { L soln }}}=0.0075 \mathrm{~mol} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \end{aligned} [/latex]

You now assume that all of the hydronium ion added [latex](0.00095[/latex] mol) reacts with acetate ion. Therefore, [latex]0.00095 \mathrm{~mol}[/latex] of acetic acid is produced and 0.00095 mol of acetate ion is used up. Hence, after reaction you have

[latex] ext { Moles of acetate ion }=(0.015-0.00095) \mathrm{mol} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-}=0.014 \mathrm{~mol} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-} [/latex]

Moles of acetic acid [latex]=(0.0075+0.00095)[/latex] mol [latex]\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}=0.0085 \mathrm{~mol} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}[/latex]

EQUILIBRIUM CALCULATION You first calculate the concentrations of [latex]\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}[/latex] and [latex]\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-}[/latex]present in the solution before you consider the acid-ionization equilibrium. Note that the total volume of solution (buffer plus hydrochloric acid) is [latex]75 \mathrm{~mL}+9.5 \mathrm{~mL}[/latex], or [latex]85 \mathrm{~mL}(0.085 \mathrm{~L})[/latex]. Hence, the starting concentrations are

[latex] \begin{gathered} {\left[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} ight]=\frac{0.0085 \mathrm{~mol}}{0.085 \mathrm{~L}}=0.10 \mathrm{M}} \ {\left[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}{ }^{-} ight]=\frac{0.014 \mathrm{~mol}}{0.085 \mathrm{~L}}=0.16 \mathrm{M}} \end{gathered} [/latex]

From this, you construct the following table:

[latex] \begin{array}{cccc} ext { Concentration }(M) & \mathrm{HC}_{2} \mathbf{H}_{3} \mathbf{O}_{2}(\mathbf{a q})+\mathrm{H}_{2} \mathbf{O}(l)& ightleftharpoons \mathrm{H}_{3} \mathbf{O}^{+}(a q)&+\mathrm{C}_{2} \mathrm{H}_{3} \mathbf{O}_{2}{ }^{-}(a q) \ ext { Starting } & 0.10 & \sim 0 & 0.16 \ ext { Change } & -x & +x & +x \ ext { Equilibrium } & 0.10-x & x & 0.16+x \end{array} [/latex]

The equilibrium-constant equation is

[latex]\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+} ight]\left[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-} ight]}{\left[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} ight]}=K_{a}[/latex]

Substituting, you get

[latex]\frac{x(0.16+x)}{0.10-x}=1.7 imes 10^{-5}[/latex]

If you assume that [latex]x[/latex] is small enough that [latex]0.16+x \simeq 0.16[/latex] and [latex]0.10-x \simeq 0.10[/latex], this equation becomes

[latex]\frac{x(0.16)}{0.10}=1.7 imes 10^{-5}[/latex]

or

[latex]x=1.7 imes 10^{-5} imes \frac{0.10}{0.16}=1.1 imes 10^{-5}[/latex]

Note that [latex]x[/latex] is indeed small, so the assumptions you made earlier are correct. The [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex]concentration is [latex]1.0 imes 10^{-5} \mathrm{M}[/latex]. The [latex]\mathrm{pH}[/latex] is

[latex]\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+} ight]=-\log \left(1.1 imes 10^{-5} ight)=\mathbf{4 . 9 6}[/latex]

Because the [latex]\mathrm{pH}[/latex] of the buffer was 5.07 (see Example 17.10), the [latex]\mathrm{pH}[/latex] has changed by [latex]5.07-4.96=0.11[/latex] units.

ADDING HCl TO PURE WATER If [latex]9.5 \mathrm{~mL}[/latex] of [latex]0.10 \mathrm{M}[/latex] hydrochloric acid were added to [latex]75 \mathrm{~mL}[/latex] of pure water, the hydronium-ion concentration would change to

[latex] \begin{aligned} {\left[\mathrm{H}_{3} \mathrm{O}^{+} ight] } & =\frac{ ext { amount of } \mathrm{H}_{3} \mathrm{O}^{+} ext {added }}{ ext { total volume of solution }} \ {\left[\mathrm{H}_{3} \mathrm{O}^{+} ight] } & =\frac{0.00095 \mathrm{~mol} \mathrm{H}_{3} \mathrm{O}^{+}}{0.085 \mathrm{~L} ext { solution }}=0.011 \mathrm{M} \end{aligned} [/latex]

(The total volume is [latex]75 \mathrm{~mL}[/latex] of water plus [latex]9.5 \mathrm{~mL} \mathrm{HCl}[/latex], assuming no change of volume on mixing.) The [latex]\mathrm{pH}[/latex] is

[latex]\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+} ight]=-\log (0.011)=1.96[/latex]

The [latex]\mathrm{pH}[/latex] of pure water is 7.00 , so the change in [latex]\mathrm{pH}[/latex] is [latex]7.00-1.96=5.04[/latex] units, compared with 0.11 units for the buffered solution.

Question 17.12: Calculating the pH of a Buffer When a Strong Acid or Strong ...

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