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Chapter 14: Problem 79

Starting with the relation $$\left[\mathrm{H}^{+}\right]=K_{\mathrm{a}} \frac{[\mathrm{HB}]}{\left[\mathrm{B}^{-}\right]}$$ derive the Henderson-Hasselbalch equation $$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log _{10} \frac{\left[\mathrm{B}^{-}\right]}{[\mathrm{HB}]}$$

Short Answer

Expert verified
Question: Derive the Henderson-Hasselbalch equation from the given relation between the concentration of protons, the ionization constant, and the concentrations of the conjugate acid and base. Answer: The Henderson-Hasselbalch equation is derived as follows: $$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log _{10} \frac{\left[\mathrm{B}^{-}\right]}{[\mathrm{HB}]}$$

Step by step solution

01

Write down the given relation

We are given the following relation: $$\left[\mathrm{H}^{+}\right]=K_{\mathrm{a}} \frac{[\mathrm{HB}]}{\left[\mathrm{B}^{-}\right]}$$
02

Take the logarithm of both sides

To derive the Henderson-Hasselbalch equation, we need to take the logarithm base 10 of both sides of the equation: $$\log_{10}\left(\left[\mathrm{H}^{+}\right]\right) = \log_{10}\left(K_{\mathrm{a}}\frac{[\mathrm{HB}]}{\left[\mathrm{B}^{-}\right]}\right)$$
03

Apply logarithm properties

Next, we apply the logarithm properties to simplify the equation. We can break down the logarithm on the right side into the difference of the logarithms of the numerator and denominator, and then we can separate the logarithm of the product into a sum of the logarithms: $$\log_{10}\left(\left[\mathrm{H}^{+}\right]\right) = \log_{10} K_{\mathrm{a}}+\log_{10} \left(\frac{[\mathrm{HB}]}{\left[\mathrm{B}^{-}\right]}\right)$$
04

Use definitions of pH and pKa

Now, we'll use the definitions of pH and pKa. Recall that pH = \(-\log_{10}\left(\left[\mathrm{H}^{+}\right]\right)\) and pKa = \(-\log_{10} K_{\mathrm{a}}\). We can rewrite the equation in terms of pH and pKa: $$- \mathrm{pH} = - \mathrm{p} K_{\mathrm{a}}+\log_{10} \left(\frac{[\mathrm{HB}]}{\left[\mathrm{B}^{-}\right]}\right)$$
05

Derive the Henderson-Hasselbalch equation

Finally, we move the pKa term to the left side of the equation to obtain the Henderson-Hasselbalch equation: $$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log _{10} \frac{\left[\mathrm{B}^{-}\right]}{[\mathrm{HB}]}$$

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Most popular questions from this chapter

A sodium hydrogen carbonate-sodium carbonate buffer is to be prepared with a \(\mathrm{pH}\) of \(9.40\). (a) What must the \(\left[\mathrm{HCO}_{3}^{-}\right] /\left[\mathrm{CO}_{3}^{2-}\right]\) ratio be? (b) How many moles of sodium hydrogen carbonate must be added to a liter of \(0.225 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) to give this \(\mathrm{pH}\) ? (c) How many grams of sodium carbonate must be added to \(475 \mathrm{~mL}\) of \(0.336 \mathrm{M} \mathrm{NaHCO}_{3}\) to give this \(\mathrm{pH}\) ? (Assume no volume change.) (d) What volume of \(0.200 \mathrm{M} \mathrm{NaHCO}_{3}\) must be added to \(735 \mathrm{~mL}\) of a \(0.139 \mathrm{M}\) solution of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to give this \(\mathrm{pH}\) ? (Assume that volumes are additive.)

A buffer is prepared by dissolving \(0.0250 \mathrm{~mol}\) of sodium nitrite, \(\mathrm{NaNO}_{2}\), in \(250.0 \mathrm{~mL}\) of \(0.0410 \mathrm{M}\) nitrous acid, \(\mathrm{HNO}_{2}\). Assume no volume change after \(\mathrm{HNO}_{2}\) is dissolved. Calculate the \(\mathrm{pH}\) of this buffer.

A buffer is prepared by dissolving \(0.037\) mol of potassium fluoride in \(135 \mathrm{~mL}\) of \(0.0237 \mathrm{M}\) hydrofluoric acid. Assume no volume change after KF is dissolved. Calculate the \(\mathrm{pH}\) of this buffer.

For an aqueous solution of acetic acid to be called "distilled white vinegar" it must contain \(5.0 \%\) acetic acid by mass. A solution with a density of \(1.05 \mathrm{~g} / \mathrm{mL}\) has a \(\mathrm{pH}\) of \(2.95 .\) Can the solution be called "distilled white vinegar"?

Explain why it is not possible to prepare a buffer with a pH of \(6.50\) by mixing \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4} \mathrm{Cl}\).
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