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Choose the correct statement from the following: (a) \(\mathrm{pOH}=-\mathrm{p} K_{b}+\log \frac{[\mathrm{Salt}]}{[\text { Base }]}\) (b) \(\mathrm{pOH}=\mathrm{p} K_{b}+\log \frac{[\mathrm{Base}]}{[\mathrm{Salt}]}\) (c) \(\mathrm{pOH}=-\mathrm{p} K_{b}-\log \frac{[\text { Salt }]}{[\text { Base }]}\) (d) \(\mathrm{pOH}=\mathrm{p} K_{b}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Base}]}\)

Short Answer

Expert verified
(d) pOH = pKb + log([Salt]/[Base])

Step by step solution

01

Analyze the structure of the equations

All the options are variants of the pOH calculation in the form of pOH= pKb+log(Base/Salt) where: pOH is the measure of base strength, pKb is the measure of the base's dissasociation providing OH-, Base is the concentration of the base and Salt is the concentration of its conjugate acid.
02

Remember the correct form of the equation

Looking at all four statements, you will notice that they differ in the sign of the equation and the ratio of the concentrations of base and salt, it's clear that the correct equation of PoH, for a basic buffer solution would be PoH= pKb + log([Salt]/[Base]). This is derived from Henderson-Hasselbalch equation with pOH rather than pKa, reflecting a base-salt scenario rather than an acid-salt scenario.
03

Choose the corresponding option

Comparing this equation with the options given in the exercise, the correct equation is found in option (d). So option (d) is the correct choice.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a formula that describes the relationship between the pH (or pOH) of a solution and the concentrations of an acid and its conjugate base (or a base and its conjugate acid). It is a vital equation for understanding acid-base buffers, because it allows us to calculate the pH of a buffer solution based on these concentrations.

In its basic form for acids, the equation is: \[ pH = pK_a + \text{log} \frac{{[\text{conjugate base}]}}{{[\text{acid}]}} \]
For bases, as seen in the textbook exercise, the equation is modified to give the pOH of the solution:\[ pOH = pK_b + \text{log} \frac{{[\text{salt}]}}{{[\text{base}]}} \]
This equation makes use of logarithmic concentration ratios to relate the known quantities (\text{pKa} or \text{pKb}) and the concentrations of the conjugate pairs to the pH or pOH.
Base Dissociation Constant (pKb)
Understanding the base dissociation constant, or pKb, is essential when working with bases and their reactions. The pKb is a logarithmic measure of the dissociation strength of a base. It indicates how readily a base will dissociate to form its conjugate acid and hydroxide (OH-) ions.

The lower the pKb value, the stronger the base. This is because a strong base dissociates more completely in solution, which corresponds to a larger dissociation constant (Kb). The pKb is calculated as the negative logarithm (to base 10) of the Kb:\[ pKb = -\text{log}(Kb) \]
Having a grasp of pKb values helps to predict the behavior of bases in acid-base reactions and plays a crucial role in buffer solutions.
Acid-Base Buffers
Acid-base buffers are solutions that resist changes in pH when small amounts of acid or base are added. They consist of a weak acid and its conjugate base or a weak base and its conjugate acid. Buffers work because the weak acid/base and its conjugate form a conjugate pair that can consume added H+ or OH- ions without significantly altering the pH.

When you add hydroxide ions (OH-) to a buffer, the weak acid present will react with them to form water and its conjugate base. When you add protons (H+), the conjugate base will consume them to form the weak acid again. This dynamic equilibrium is what maintains the solution's pH at a nearly constant level.
Logarithmic Concentration Ratios
Logarithmic concentration ratios are a tool used in chemistry to examine the relationship between concentrations of two substances in a solution. This approach is used in the Henderson-Hasselbalch equation. The ratio of the concentrations of the conjugate acid/base pair is vital to determine the pH or pOH of a solution.

The logarithmic nature of these ratios allows us to easily calculate changes in pH or pOH because a logarithmic scale is more manageable when dealing with the wide range of hydrogen ion concentrations found in solutions. This log ratio also explains why a small change in concentration can result in a significant change in pH, critical for understanding buffer systems.

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Most popular questions from this chapter

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 an acid buffer is a solution of (a) ammonium acetate (b) a mixture of \(\mathrm{NH}_{4} \mathrm{OH}+\mathrm{NH}_{4} \mathrm{Cl}\) (c) a mixture of \(\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{CH}_{3} \mathrm{COONa}\) (d) a solution of \(\mathrm{CH}_{3} \mathrm{COONa}\)

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) \(-p k_{a}+\log \frac{[\text { Salt }]}{[\text { Acid }]}\)

Both anionic and cationic hydrolysis occur in the case of (a) salts of strong acid and strong base (b) salts of strong acid and weak base (c) salts of weak acid and strong base (d) salts of weak acid and weak base

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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