Calculating the pH of a Buffer from Given Volumes of Solution Instructions for making up a buffer say to mix 60. mL (0.060 L) of 0.100 M NH3 with 40. mL (0.040 L) of 0.100 M NH4Cl. What is the pH of this buffer? PROBLEM STRATEGY The buffer contains a base and its conjugate acid in equilibrium. The

Question

Calculating the pH of a Buffer from Given Volumes of Solution

Instructions for making up a buffer say to mix [latex]60 . \mathrm{mL}(0.060 \mathrm{~L})[/latex] of [latex]0.100 M \mathrm{NH}_{3}[/latex] with 40. mL (0.040 L) of 0.100 M [latex]NH_4Cl[/latex]. What is the pH of this buffer?

PROBLEM STRATEGY

The buffer contains a base and its conjugate acid in equilibrium. The equation is

[latex]\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q)[/latex]

The problem is to obtain the concentration of [latex]\mathrm{H}_{3} \mathrm{O}^{+}[/latex]or [latex]\mathrm{OH}^{-}[/latex]in this equilibrium mixture. To do the equilibrium calculation, you need the starting concentrations in the solution obtained by mixing the [latex]\mathrm{NH}_{3}[/latex] and [latex]\mathrm{NH}_{4} \mathrm{Cl}[/latex] solutions. For this, you calculate the moles of [latex]\mathrm{NH}_{3}[/latex] and moles of [latex]\mathrm{NH}_{4}{ }^{+}[/latex]added to the buffer solution, then divide by the total volume of buffer. Then you are ready to do the equilibrium calculation.

Accepted Answer

How many moles of [latex]\mathrm{NH}_{3}[/latex] are added? Recall that

[latex] ext { Molarity } \mathrm{NH}_{3}=\frac{ ext { moles } \mathrm{NH}_{3}}{ ext { liters } \mathrm{NH}_{3} ext { solution }}[/latex]

Note that the instructions say to add [latex]60 . \mathrm{mL}[/latex] (or [latex]0.060 \mathrm{~L}[/latex] ) of [latex]0.100 \mathrm{M} \mathrm{NH}_{3}[/latex]. So

[latex] \begin{aligned} ext { Moles }\,\mathrm{NH}_{3} & = ext { molarity } \mathrm{NH}_{3} imes ext { liters } \mathrm{NH}_{3} ext { solution } \ & =0.100 \frac{\mathrm{mol}}{\mathrm{L}} imes 0.060 \mathrm{~L}=0.0060 \mathrm{~mol} \mathrm{NH}_{3} \end{aligned} [/latex]

In the same way, you find that you have added [latex]0.0040 \mathrm{~mol} \mathrm{NH}_{4}^{+}[/latex](from [latex]\mathrm{NH}_{4} \mathrm{Cl}[/latex] ). Assume that the total volume of buffer equals the sum of the volumes of the two solutions.

[latex] ext { Total volume buffer }=60 . \mathrm{mL}+40 . \mathrm{mL}=100 . \mathrm{mL}(0.100 \mathrm{~L})[/latex]

Therefore, the concentrations of base and conjugate acid are

[latex] \begin{aligned} {\left[\mathrm{NH}_{3} ight] } & =\frac{0.0060 \mathrm{~mol}}{0.100 \mathrm{~L}}=0.060 \mathrm{M} \ {\left[\mathrm{NH}_{4}{ }^{+} ight] } & =\frac{0.0040 \mathrm{~mol}}{0.100 \mathrm{~L}}=0.040 \mathrm{M} \end{aligned} [/latex]

STEP 1 Fill in the concentration table for the acid-base equilibrium (base ionization of [latex]\mathrm{NH}_{3}[/latex] ).

[latex]\begin{array}{lccc} ext { Concentration }(M) & \mathbf{N H}_{3}(\boldsymbol{a q})+\mathbf{H}_{\mathbf{2}} \mathbf{O}(\boldsymbol{l}) & ightleftharpoons \mathbf{N H}_{\mathbf{4}}{ }^{+}(\boldsymbol{a q})&+\mathbf{O H}^{-}(\boldsymbol{a q}) \ ext { Starting } & 0.060 & 0.040 & 0 \ ext { Change } & -x & +x & +x \ ext { Equilibrium } & 0.060-x & 0.040+x & x\end{array}[/latex]

STEP 2 You substitute the equilibrium concentrations into the equilibrium-constant equation.

[latex] \begin{aligned} \frac{\left[\mathrm{NH}_{4}{ }^{+} ight]\left[\mathrm{OH}^{-} ight]}{\left[\mathrm{NH}_{3} ight]} & =K_{b} \ \frac{(0.040+x) x}{0.060-x} & =1.8 imes 10^{-5} \end{aligned} [/latex]

STEP 3 To solve this equation, you assume that [latex]x[/latex] is small compared with 0.040 and 0.060 . So this equation becomes

[latex]\frac{0.040 x}{0.060}=1.8 imes 10^{-5}[/latex]

Therefore, [latex]x=(0.060 / 0.040) imes\left(1.8 imes 10^{-5} ight)=2.7 imes 10^{-5}[/latex]. (Check that [latex]x[/latex] can be neglected in [latex]0.040+x[/latex] and [latex]0.060-x[/latex].) Thus, the hydroxide-ion concentration is [latex]2.7 imes 10^{-5} M[/latex]. The [latex]\mathrm{pH}[/latex] of the buffer is

[latex]\mathrm{pH}=14.00-\mathrm{pOH}=14.00+\log \left(2.7 imes 10^{-5} ight)=\mathbf{9 . 4 3}[/latex]

(If you have [latex]2.7 imes 10^{-5}[/latex] in your calculator from the previous calculation, you obtain the [latex]\mathrm{pH}[/latex] by pressing the [latex]\log[/latex] button, then adding 14.)

Question 17.11: Calculating the pH of a Buffer from Given Volumes of Solutio...

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