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Chapter 17: Problem 9

Calculate the \(\mathrm{pH}\) of a buffer that is (a) \(0.012 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\left(K_{\mathrm{a}}=6.3 \times 10^{-5}\right)\) and 0.033 \(\mathrm{M} \mathrm{NaC}_{6} \mathrm{H}_{5} \mathrm{COO}\) (b) \(0.408 \mathrm{M} \mathrm{NH}_{3}\) and \(0.153 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\)

Short Answer

Expert verified
The pH of a buffer solution for part (a) is 5.02, and for part (b) is 3.93.

Step by step solution

01

Identify the buffer components and understand the Henderson-Hasselbalch equation

The buffer system contains a weak acid C6H5COOH and its conjugate base NaC6H5COO in part (a), and a weak base NH3 and its conjugate acid NH4Cl in part (b). The Henderson–Hasselbalch equation is as follows: \[pH = pKa + log_{10} \left(\frac{[A^-]}{[HA]}\right)\] for the acid and its conjugate base, and \[pOH = pKb + log_{10} \left(\frac{[B]}{[HB^+]}\right)\] for the base and its conjugate acid, where [A^-] represents the concentration of the base, [HA] the concentration of the acid, [B] the concentration of the base, and [HB^+] the concentration of the conjugate acid. We will determine pH using these equations.
02

Calculate the pH for part (a)

Substitute the given values into the Henderson-Hasselbalch equation for the acid and its conjugate base: \[ Ka = 6.3 × 10^{−5}\], \[ [A^-] = [NaC6H5COO] = 0.033 M\], and \[ [HA] = [C6H5COOH] = 0.012 M\]. First, calculate pKa = -log(Ka) = -log(6.3 × 10^{−5}) = 4.2, then substitute these values into the equation, we get: pH = pKa + log_{10} \left(\frac{[A^-]}{[HA]}\right) = 4.2 + log_{10} \left(\frac{0.033}{0.012}\right) = 4.2 + 0.82 = 5.02.
03

Calculate the pH for part (b)

Now use the Henderson-Hasselbalch equation for a base and its conjugate acid. The base is NH3 with a given Kb of 1.8 × 10^{-5}. For NH3 and NH4Cl, \[ [B] = [NH3] = 0.408 M\] and \[ [HB^+] = [NH4Cl] = 0.153 M\]. Start by converting Kb to Ka using the equation \[ Ka = \frac{Kw}{Kb} = \frac{1.0 × 10^{−14}}{1.8 × 10^{−5}} = 5.56 × 10^{−10}\]. Then, calculate pKa = -log(Ka) = -log(5.56 × 10^{−10}) = 9.25. Now, find pOH using the Henderson-Hasselbalch equation: \[pOH = pKa + log_{10} \left(\frac{[B]}{[HB^+]}\right) = 9.25 + log_{10} \left(\frac{0.408}{0.153}\right) = 9.25 + 0.82 = 10.07\]. Convert pOH to pH using the equation \[pH = 14 - pOH = 14 - 10.07 = 3.93.\]
04

Interpret the results

The pH of a buffer solution for part (a) is 5.02, and for part (b) is 3.93. These measurements indicate that both buffers are acidic since their pH values are less than 7. pH of part (b) is lower than that of part (a) which means it is more acidic.

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Most popular questions from this chapter

What is the \(\mathrm{pH}\) of a solution obtained by adding \(1.15 \mathrm{mg}\) of aniline hydrochloride \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \mathrm{Cl}^{-}\right)\) to \(3.18 \mathrm{L}\) of \(0.105 \mathrm{M}\) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right) ?\)

To convert \(\mathrm{NH}_{4}^{+}(\text {aq })\) to \(\mathrm{NH}_{3}(\mathrm{aq}),\) (a) add \(\mathrm{H}_{3} \mathrm{O}^{+}\) (b) raise the \(\mathrm{pH} ;\) (c) add \(\mathrm{KNO}_{3}(\mathrm{aq}) ;\) (d) add \(\mathrm{NaCl}\).

Briefly describe each of the following ideas, phenomena, or methods: (a) the common-ion effect; (b) the use of a buffer solution to maintain a constant \(\mathrm{pH}\) (c) the determination of \(\mathrm{p} K_{\mathrm{a}}\) of a weak acid from a titration curve; (d) the measurement of \(\mathrm{pH}\) with an acid-base indicator.

Even though the carbonic acid-hydrogen carbonate buffer system is crucial to the maintenance of the \(\mathrm{pH}\) of blood, it has no practical use as a laboratory buffer solution. Can you think of a reason(s) for this?

In the titration of \(10.00 \mathrm{mL}\) of \(0.04050 \mathrm{M} \mathrm{HCl}\) with \(0.01120 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) in the presence of the indicator 2,4-dinitrophenol, the solution changes from colorless to yellow when 17.90 mL of the base has been added. What is the approximate value of \(\mathrm{p} K_{\mathrm{HIn}}\) for 2,4 -dinitrophenol? Is this a good indicator for the titration?
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