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Chapter 17: Problem 4

The \(\mathrm{pH}\) of an acidic buffer according to the Henderson equation is given by (a) \(\mathrm{p} K_{a}-\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}\) (b) \(\mathrm{pK}_{a}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}\) (c) \(\mathrm{p} \boldsymbol{K}_{a}+\log \frac{[\mathrm{Acid}]}{[\mathrm{Salt}]}\) (d) \(-p k_{a}+\log \frac{[\text { Salt }]}{[\text { Acid }]}\)

Short Answer

Expert verified
The correct formulation of the Henderson equation that represents the pH of an acidic buffer solution is (b) \(\mathrm{pK}_{a}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}\).

Step by step solution

01

Understanding the Henderson Equation

The Henderson-Hasselbalch equation is a simplified expression of the law of mass action which is used to calculate the pH of buffer solutions. The equation is given as \(pH = pKa + \log \frac{[Salt]}{[Acid]}\) .
02

Analyzing the Options

We will analyze each option and compare it with the original Henderson equation. (a) \(\mathrm{p} K_{a}-\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}\): In this option, minus sign appears instead of plus in the equation which is incorrect. In options (c) and (d), the expressions after the logarithm are reversed or incorrectly sign as compared to the actual equation and hence they are incorrect. However, option (b) \(\mathrm{pK}_{a}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Acid}]}\) perfectly matches to the actual Henderson equation and hence, it is the correct formulation.

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Most popular questions from this chapter

In spite of various acid and base-producing reactions in our body, the pH of our blood is (a) \(5.4\) (b) \(6.4\) (c) \(7.4\) (d) \(8.4\)

Heat of neutralisation of a strong acid and weak base is less than \(57.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\). This is because (a) all the acid is not ionised (b) all the base is not ionised (c) the ionisation is complete but further reaction does not take place (d) some of the heat is transferred to the surroundings

The \(\mathrm{pH}\) value of a solution obtained by mixing \(5 \mathrm{~g}\) of acetic acid and \(7.5 \mathrm{~g}\) of sodium acetate and making the volume to \(500 \mathrm{~mL}\left(\mathrm{~K}_{\mathrm{CH}_{3} \mathrm{COOH}}=1.8 \times 10^{-5}\right)\) is (a) \(4.78\) (b) \(3.78\) (c) \(5.78\) (d) zero

An example of a basic buffer is a solution of (a) \(\mathrm{NH}_{4} \mathrm{OH}+\mathrm{NH}_{4} \mathrm{Cl}\) (b) \(\mathrm{NH}_{4} \mathrm{OH}+\mathrm{HCl}\) (c) \(\mathrm{CH}_{3} \mathrm{COONH}_{4}\) (d) \(\mathrm{NH}_{4} \mathrm{OH}+\mathrm{NaOH}\)

The \(\mathrm{pH}\) of a solution of salt of strong acid and weak base is given by a) \(\frac{1}{2}\left(\log K_{w^{\prime}}+\log K_{b}+\log C\right)\) (b) \(\frac{1}{2}\left(\log K_{w}-\log K_{b}-\log C\right)\) c) \(\frac{1}{2}\left(\log K_{w^{\prime}}-\log K_{b}+\log C\right)\) d) \(\frac{1}{2}\left(-\log K_{w^{\prime}}-\log K_{b}-\log C\right)\)
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