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Chapter 17: Problem 88

The Henderson-Hasselbalch equation can be written as \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}-\log \left(\frac{1}{\alpha}-1\right)\) where \(\alpha=\frac{\left[\mathrm{A}^{-}\right]}{\left[\mathrm{A}^{-}\right]+[\mathrm{HA}]}\) Thus, the degree of ionization \((\alpha)\) of an acid can be determined if both the \(\mathrm{pH}\) of the solution and the \(\mathrm{p} K_{\mathrm{a}}\) of the acid are known. (a) Use this equation to plot the pH versus the degree of ionization for the second ionization constant of phosphoric acid \(\left(K_{\mathrm{a}}=6.3 \times 10^{-8}\right)\) (b) If \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\) what is the degree of ionization? (c) If the solution had a pH of 6.0 what would the value of \(\alpha\) be?

Short Answer

Expert verified
For the phosphoric acid with \(Ka = 6.3 \times 10^-8\), the \(pKa = 7.20\). By plotting pH against degree of ionization, it will show a negative slope and intercept at \(pKa\). When pH = pKa, the degree of ionization is 0.50. If the solution has a pH of 6.0, the degree of ionization (alpha) is 0.7937.

Step by step solution

01

Determine the pKa.

The second ionization constant of phosphoric acid is given as: \(K_a=6.3 \times 10^{-8}\). To convert \(K_a\) to \(pKa\), take the negative logarithm of the \(K_a\) (i.e. \(pKa = -\log(K_a)\)). This gives \(pKa = -\log(6.3 \times 10^{-8}) = 7.20\).
02

Express pH in terms of pKa and alpha.

The Henderson–Hasselbalch equation can be rewritten to express \(pH\) in terms of \(pKa\) and \(\alpha\). Essentially, this equation is linear with slope \(= -1\) and intercept \(= pKa\). This allows you to plot the relationship between \(pH\) and \(\alpha\).
03

Calculate Degree of Ionization When pH = pKa.

When \(pH = pKa\), substituting into Henderson-Hasselbalch equation, \(-\log(\frac{1}{\alpha} - 1) = 0\). Solving for alpha, we will get: \(\alpha = \frac{1}{2}\). This degree of ionization is the value of alpha when \(pH = pKa\).
04

Calculate Degree of Ionization for a Given pH.

Given pH of 6.0, we substitute \(pH = 6.0\) and \(pKa = 7.20\) into the Henderson-Hasselbalch equation. Solving for \(\alpha\), we will get \(\alpha\) as the degree of ionization for a \(pH\) of 6.0. \(pH = 7.20 - \log\left(\frac{1}{\alpha} - 1\right)\) can be rearranged as: \(\alpha = \frac{1}{10^{(7.20 - 6.0)} + 1}\), hence \(\alpha = \frac{1}{1.26} = 0.7937\).

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Most popular questions from this chapter

Lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH},\) is found in sour milk. A solution containing \(1.00 \mathrm{g} \mathrm{NaCH}_{3} \mathrm{CH}_{2} \mathrm{COO}\) in 100.0 \(\mathrm{mL}\) of \(0.0500 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\) has a \(\mathrm{pH}=4.11 .\) What is \(K_{\mathrm{a}}\) of lactic acid?

Calculate the \(\mathrm{pH}\) at the points in the titration of \(25.00 \mathrm{mL}\) of \(0.132 \mathrm{M} \mathrm{HNO}_{2}\) when (a) \(10.00 \mathrm{mL}\) and (b) 20.00 mL of 0.116 M \(\mathrm{HNO}_{2}\) when \((\mathrm{a}) 10.00 \mathrm{mL}\) and (b) \(20.00 \mathrm{mL}\) of \(0.116 \mathrm{M}\) NaOH have been added. For \(\mathrm{HNO}_{2}, K_{\mathrm{a}}=7.2 \times 10^{-4}\) \(\mathrm{HNO}_{2}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NO}_{2}^{-}\)

Solution (a) is \(100.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) and solution (b) is \(150.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{NaCH}_{3} \mathrm{COO}\). A few drops of thymol blue indicator are added to each solution. What is the color of each solution? What is the color of the solution obtained when these two solutions are mixed?

An acetic acid-sodium acetate buffer can be prepared by the reaction \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O}\) (From \(\mathrm{NaCH}_{3} \mathrm{COO}\) )(From HCl) (a) If \(12.0 \mathrm{g} \mathrm{NaCH}_{3} \mathrm{COO}\) is added to \(0.300 \mathrm{L}\) of 0.200 M HCl, what is the pH of the resulting solution? (b) If \(1.00 \mathrm{g} \mathrm{Ba}(\mathrm{OH})_{2}\) is added to the solution in part (a), what is the new pH? (c) What is the maximum mass of \(\mathrm{Ba}(\mathrm{OH})_{2}\) that can be neutralized by the buffer solution of part (a)? (d) What is the pH of the solution in part (a) following the addition of \(5.50 \mathrm{g} \mathrm{Ba}(\mathrm{OH})_{2} ?\)

Calculate \(\left[\mathrm{OH}^{-}\right]\) in a solution that is (a) \(0.0062 \mathrm{M}\) \(\mathrm{Ba}(\mathrm{OH})_{2}\) and \(0.0105 \mathrm{M} \mathrm{BaCl}_{2} ;\) (b) \(0.315 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) and \(0.486 \mathrm{M} \mathrm{NH}_{3} ;\) (c) \(0.196 \mathrm{M} \mathrm{NaOH}\) and \(0.264 \mathrm{M}\) \(\mathrm{NH}_{4} \mathrm{Cl}\)
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