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The \(\mathrm{pH}\) at the equivalence point when a solution of \(0.01 \mathrm{M}-\mathrm{CH}_{3} \mathrm{COOH}\) is titrated with a solution of \(0.01 \mathrm{M}-\mathrm{NaOH}\), is \(\left(\mathrm{p} K_{\mathrm{a}}\right.\) of \(\mathrm{CH}_{3} \mathrm{COOH}=4.7, \log 5=0.7\) ) (a) \(10.5\) (b) \(3.5\) (c) \(10.35\) (d) \(3.65\)

Short Answer

Expert verified
The pH at the equivalence point is 9.25.

Step by step solution

01

Understanding the Situation

Recognize that we're dealing with a weak acid (acetic acid, \(\mathrm{CH}_{3}\mathrm{COOH}\)) being titrated with a strong base (sodium hydroxide, \(\mathrm{NaOH}\)). At the equivalence point, the amount of acid equals the amount of base, resulting in the formation of the conjugate base of acetic acid (acetate ion, \(\mathrm{CH}_{3}\mathrm{COO}^{-}\)). The \(\mathrm{pH}\) at this point depends on the hydrolysis of the acetate ion.
02

Calculating the Concentration of Acetate Ion

Since the molarity of \(\mathrm{CH}_{3}\mathrm{COOH}\) is the same as the molarity of \(\mathrm{NaOH}\), all the acetic acid will be converted into acetate ions. Therefore, the concentration of acetate ions at the equivalence point is also \(0.01\,\mathrm{M}\).
03

Applying the Hydrolysis Equation

The acetate ion hydrolyzes in water to form acetic acid and hydroxide ions. The equilibrium expression for its hydrolysis is: \[\mathrm{CH}_{3}\mathrm{COO}^{-} + H_2O \rightleftharpoons \mathrm{CH}_{3}\mathrm{COOH} + OH^{-}\]. The \(\mathrm{K}_{b}\) of the acetate ion can be calculated from the given \(\mathrm{pK}_{a}\) using the formula \(\mathrm{K}_{w} = \mathrm{K}_{a} \times \mathrm{K}_{b}\), where \(\mathrm{K}_{w}\) is the ion product of water (which is \(1 \times 10^{-14}\) at 25 degrees Celsius).
04

Calculating the \(\mathbf{K}_{b}\) Value

First, find \(\mathrm{K}_{a}\) from \(\mathrm{pK}_{a}\) using the equation \(\mathrm{K}_{a} = 10^{-\mathrm{pK}_{a}}\). Then calculate \(\mathrm{K}_{b}\) using \(\mathrm{K}_{w} = \mathrm{K}_{a} \times \mathrm{K}_{b}\) and the relationship \(\mathrm{K}_{b} = \frac{\mathrm{K}_{w}}{\mathrm{K}_{a}}\).
05

Calculating Hydroxide Ion Concentration

Assuming the hydrolysis of acetate ions is a weak reaction, use the approximated equilibrium concentration \([OH^{-}] = \sqrt{\mathrm{K}_{b} \times [\mathrm{CH}_{3}\mathrm{COO}^{-}]}.\)
06

Finding \(\mathrm{pOH}\)

Use the concentration of hydroxide ions to find \(\mathrm{pOH}\) using the formula \(\mathrm{pOH} = -\log[OH^{-}]\).
07

Calculating the \(\mathrm{pH}\)

Use the relationship between \(\mathrm{pH}\) and \(\mathrm{pOH}\) in pure water, which is \(\mathrm{pH} + \mathrm{pOH} = 14\), to find the \(\mathrm{pH}\) at the equivalence point.

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Key Concepts

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

Acid-Base Titration
Acid-base titration is a laboratory procedure used to determine the concentration of an acid or base in a solution. During the titration, an acid is neutralized by a base, or vice versa, to reach the equivalence point where the number of moles of hydrogen ions equals the number of moles of hydroxide ions. The process typically involves a pH indicator or a pH meter to detect when the equivalence point is achieved. An important characteristic of titration is the point of neutrality that differs based on the nature of the substances involved; in the case of a weak acid and strong base titration, the equivalence point pH is greater than 7 due to the resulting conjugal base's hydrolysis. This is essential for understanding the behavior of the acetate ion in our problem.
Hydrolysis of Acetate Ion
Hydrolysis refers to the reaction of a salt with water to form an acid and a base. When acetic acid, a weak acid, is titrated with a strong base like sodium hydroxide, the products are water and the sodium acetate salt. Sodium acetate dissociates in water to produce acetate ions, which then undergo hydrolysis to form hydroxide ions and acetic acid:
\[CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-\]
The generation of hydroxide ions increases the solution's pH, making it basic at the equivalence point. The extent of hydrolysis and the concentration of hydroxide ions can be predicted using equilibrium constants, which brings us to the importance of the pKa and Kb relationship.
pKa and Kb Relationship
The pKa value is a measure of the acidity of a substance, the lower the pKa value, the stronger the acid. Conversely, Kb is the base dissociation constant and provides information on the strength of a base - the higher the value, the stronger the base. There's a quantitative relationship between pKa and Kb, given by the water ion product, Kw, which is \(1 \times 10^{-14}\) at 25 degrees Celsius:
\[K_w = K_a \times K_b\]
For a conjugate acid-base pair, knowing one of the constants allows you to calculate the other. Since acetate is the conjugate base of acetic acid, we can calculate its Kb if we know the pKa of acetic acid and the value of Kw. This relationship is incredibly useful for predicting the behavior of solutions in acid-base reactions.
Equilibrium Concentration Calculation
Equilibrium concentration calculations are essential for predicting the extent of reactions in solutions. In the context of our titration problem, once Kb is known, it can be used to calculate the concentration of hydroxide ions in solution at the equivalence point. The formula is:
\[ [OH^-] = \sqrt{K_b \times [CH_3COO^-]} \]
Given that the initial concentration of acetate ions comes directly from the titrated weak acid's concentration, we can calculate \([OH^-]\) and thus the pOH. Once pOH is known, subtracting it from 14 gives us the pH at the equivalence point. This series of calculations allows students to predict the pH of the solution after complete neutralization in a titration experiment.

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Most popular questions from this chapter

To a solution of acetic acid, solid sodium acetate is gradually added. When ' \(\mathrm{x} \mathrm{g}\) ' of the salt has been added, the \(\mathrm{pH}\) has a certain value. When total 'y g' of the salt has been added, the \(\mathrm{pH}\) has been further raised by \(0.6\) units. What is the ratio of \(x: y ?(\log 3.98=0.6)\) (a) \(3.98: 1\) (b) \(1: 3.98\) (c) \(2: 3.98\) (d) \(3.98: 2\)

Calculate pH of \(0.02 \mathrm{M}-\) HA solution. \(K_{\mathrm{a}}\) for \(\mathrm{HA}=2 \times 10^{-12} .(\log 2=0.3\) \(\log 3=0.48\) ) (a) \(6.65\) (b) \(6.70\) (c) \(6.85\) (d) \(6.52\)

\(\begin{array}{ll}\text { The } \text { equilibrium } & \text { carbonate } \text { ion }\end{array}\) concentration after equal volumes of \(0.7 \mathrm{M}-\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(0.7 \mathrm{M}-\mathrm{HCl}\) solutions are mixed, is \(\left(K_{\mathrm{al}}\right.\) and \(K_{\mathrm{a} 2}\) for \(\mathrm{H}_{2} \mathrm{CO}_{3}\) are \(4.9 \times 10^{-6}\) and \(4.0 \times 10^{-11}\), respectively) (a) \(0.7 \mathrm{M}\) (b) \(0.35 \mathrm{M}\) (c) \(0.002 \mathrm{M}\) (d) \(0.001 \mathrm{M}\)

What is the solubility of \(\mathrm{MnS}\) in pure water, assuming hydrolysis of \(\mathrm{S}^{2-}\) ions? \(K_{\mathrm{sp}}\) of \(\mathrm{MnS}=2.5 \times 10^{-10}, K_{\mathrm{al}}=1 \times 10^{-7}\) and \(K_{\mathrm{a} 2}=1 \times 10^{-14}\) for \(\mathrm{H}_{2} \mathrm{~S} .\left(0.63^{3}=0.25\right)\) (a) \(6.3 \times 10^{-4} \mathrm{M}\) (b) \(2.5 \times 10^{-4} \mathrm{M}\) (c) \(6.3 \times 10^{-3} \mathrm{M}\) (d) \(1.58 \times 10^{-5} \mathrm{M}\)

The dissociation constant of acetic acid is \(0.000018\) and that for cyanoacetic acid is \(0.0036\) at \(298 \mathrm{~K}\). What would be the ratio of volumes of the two acid solutions, each containing equal moles of the acids, so that the solutions becomes isohydric? (a) \(1: 1\) (b) \(1: \sqrt{200}\) (c) \(1: 200\) (d) \(200: 1\)

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