Which of the following pairs of ions would make the best buffer with a pH between 6 and 7? \(K_{\mathrm{a}}\) for \(\mathrm{HC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}=1.75 \times 10^{-5}\) \(K_{\mathrm{a}}\) for \(\mathrm{HPO}_{4}^{2-}=4.8 \times 10^{-13}\) (A) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}\) (B) \(\mathrm{HPO}_{4}^{2-}\) and \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) (C) \(\mathrm{HC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}\) and \(\mathrm{NaC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}\) (D) \(\mathrm{NaOH}\) and \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

Short Answer

Expert verified
The most suitable buffer with a pH between 6 and 7 is created using \(\mathrm{HC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}\) and \(\mathrm{NaC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}\). Therefore, the answer is C.

Step by step solution

01

Examine all options

Look at each option and identify the conjugate acid-base pairs. For option A, the pairs are \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}\), which are not conjugate pairs. For option B, the pairs are \(\mathrm{HPO}_{4}^{2-}\) and \(\mathrm{Na}_{3} \mathrm{PO}_{4}\), where \(\mathrm{PO}_{4}^{3-}\) is the base. Option C has pairs \(\mathrm{HC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}\) and $\mathrm{NaC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}$, where \(\mathrm{C}_{3} \mathrm{H}_{2} \mathrm{O}_{2}^{-}\) is the base. Finally, option D pairs \(\mathrm{NaOH}\) and \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\), but \(\mathrm{NaOH}\) is a strong base and therefore doesn't make good buffers.
02

Check the pH range

Using Henderson-Hasselbalch equation, \(\mathrm{pH} = \mathrm{pKa} + \log_{10}\left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right)\), the best buffer will have a pKa close to the desired pH, which is between 6 and 7 in this case. From the given acid dissociation constants, calculate pKa by taking the negative logarithm of Ka. For \(\mathrm{HC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}\), pKa = -log(1.75 x 10^-5) = 4.76. For \(\mathrm{HPO}_{4}^{2-}\), pKa = -log(4.8 x 10^-13) = 12.32. It's clear that \(\mathrm{HC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}\)'s pKa is closer to the desired pH range.
03

Choose the correct answer

Based on the discussion above, option C, \(\mathrm{HC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}\) and $\mathrm{NaC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}$, makes the most suitable buffer with a pH between 6 and 7. The other options either do not form conjugate acid-base pairs or their pKa values are not within the vicinity of the desired pH range.

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