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Chapter 7: Problem 30

Which of the following is correct? (a) The \(\mathrm{pH}\) of one litre solution containing \(0.49 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is \(2.0\) (b) The conjugate base of \(\mathrm{H}_{2} \mathrm{~S}\) is \(\mathrm{S}^{2-}\). (c) \(\mathrm{BF}_{3}\) is lewis base. (d) Phenolphthalein is colourless in basic medium.

Short Answer

Expert verified
Option (a) is correct; the pH of the solution is 2.0.

Step by step solution

01

Evaluate Option (a) - Calculating pH

First, calculate the number of moles of \( \mathrm{H}_{2}\mathrm{SO}_{4} \). The molar mass of \( \mathrm{H}_{2}\mathrm{SO}_{4} \) is approximately \( 98 \; \text{g/mol} \). So, the moles present in \( 0.49 \; \text{g} \) of \( \mathrm{H}_{2}\mathrm{SO}_{4} \) is \( \frac{0.49}{98} \approx 0.005 \) moles. Since \( \mathrm{H}_{2}\mathrm{SO}_{4} \) is a strong acid and dissociates completely releasing 2 moles of \( \mathrm{H^+} \) ions per mole of acid, the concentration of \( \mathrm{H^+} \) ions is \( 2 \times 0.005 = 0.01 \; \text{mol/L} \). The \( \mathrm{pH} \) is \( -\log[\mathrm{H^+}] = -\log[0.01] = 2.0 \). Therefore, option (a) is correct.
02

Evaluate Option (b) - Conjugate Base Identification

The conjugate base of an acid is formed when the acid donates a proton \( \mathrm{H^+} \). For \( \mathrm{H}_{2}\mathrm{~S} \), the conjugate base is formed as \( \mathrm{HS}^- \) when one \( \mathrm{H}^+ \) is lost. \( \mathrm{S}^{2-} \) would be the result of losing two protons, so option (b) is not correct.
03

Evaluate Option (c) - Lewis Character of BF3

A Lewis base is defined as an electron pair donor. \( \mathrm{BF}_{3} \) is an electron-deficient compound and tends to accept electron pairs to complete its octet, thereby acting as a Lewis acid, not a base. Thus, option (c) is incorrect.
04

Evaluate Option (d) - Behavior of Phenolphthalein

Phenolphthalein changes color based on the pH of the solution. It is colorless in a neutral and acidic medium and turns pink to fuchsia in a basic medium. Therefore, option (d) is incorrect.

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Key Concepts

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

pH Calculation
Calculating the pH of a solution involves determining the concentration of hydrogen ions \( [ ext{H}^+ ] \) in the solution and applying the formula \( ext{pH} = -\log [ ext{H}^+ ] \). In the problem, you are required to find the pH of a solution containing \( 0.49 \) g of \( ext{H}_2 ext{SO}_4 \) in one liter of water. The molar mass of sulfuric acid is approximately \( 98\) g/mol, which means the solution contains \( \frac{0.49}{98} \approx 0.005 \) moles of \( ext{H}_2 ext{SO}_4 \). Since sulfuric acid is a strong acid that dissociates completely, each mole of sulfuric acid releases two moles of hydrogen ions, resulting in \( 2 \times 0.005 = 0.01 \) moles per liter. The pH is then calculated as \( -\log [0.01] = 2.0 \). This confirms that the pH is indeed 2.0, and option (a) from the exercise is correct.
Conjugate Base
Understanding the concept of conjugate bases is essential in acid-base chemistry. A conjugate base is formed when an acid loses a proton \( \text{H}^+ \). For the acid \( \text{H}_2 ext{S} \), the conjugate base is \( \text{HS}^- \), which results from the loss of one proton. The option in the exercise incorrectly suggests \( \text{S}^{2-} \) as the conjugate base of \( \text{H}_2 ext{S} \), which would occur only after losing two protons. In acid-base reactions, the original acid and its conjugate base are connected through equilibrium processes.
  • When an acid donates a proton, it forms its conjugate base.
  • Conversely, when the conjugate base accepts a proton, it reforms the acid.
This fundamental concept helps in understanding the behavior of acids and bases in various chemical reactions.
Lewis Acids and Bases
The concept of Lewis acids and bases broadens the definition of acids and bases beyond proton transfer. A Lewis acid is an electron pair acceptor, while a Lewis base is an electron pair donor. Regarding \( ext{BF}_3 \), it acts as a Lewis acid because it is electron-deficient; it tends to accept electron pairs to achieve a stable octet. The exercise mentions \( ext{BF}_3 \) as a Lewis base, which is incorrect. In fact,
  • Lewis acids are often compounds with empty orbitals capable of accepting electron pairs.
  • Lewis bases typically have lone pairs that are available for donation.
Understanding Lewis structures is essential for predicting how compounds will react, particularly in coordination chemistry and organic reactions.
Indicators and pH
Indicators are substances that change color depending on the pH of the solution. They are essential for determining the approximate pH or acidity/basicity of a solution. Phenolphthalein is a common indicator that is colorless in acidic solutions and turns pink to fuchsia as the solution becomes more basic. The statement in the exercise claiming that phenolphthalein is colorless in a basic medium is incorrect. Some important points about phenolphthalein:
  • In acidic (pH < 7) and neutral (pH = 7) solutions, phenolphthalein remains colorless.
  • In basic solutions (pH > 8.2), it changes to pink/fuchsia indicating an increase in pH.
Understanding the use of pH indicators like phenolphthalein is crucial in titrations where precise determination of the endpoint is necessary.

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Most popular questions from this chapter

In which of the following gaseous reaction, \(\mathrm{K}_{\mathrm{p}}\) and \(\mathrm{K}_{\mathrm{c}}\) have the same values: (a) \(2 \mathrm{Hl} \rightleftharpoons \mathrm{H}_{2}+\mathrm{I}_{2}\) (b) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) (c) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) (d) \(\mathrm{PCI}_{5} \rightleftharpoons \mathrm{PCI}_{3}+\mathrm{Cl}_{2}^{3}\)

In which of the following gaseous reaction, the value of \(K_{p}\) is less than \(K_{c}:\) (a) \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCI}_{3}+\mathrm{Cl}_{2}\) (b) \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) (c) \(2 \mathrm{HI} \rightleftharpoons \mathrm{H}_{2}+\mathrm{I}_{2}\) (d) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\)

The reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~s})+\mathrm{Cl}_{2}(\mathrm{~g})\) is an example of: (a) Backward reaction (b) Forward reaction (c) Irreversible reaction (d) Reversible reaction

One mole of a compound AB reacts with one mole of compound CD according to the equation: \(\mathrm{AB}+\mathrm{CD} \rightarrow \mathrm{AD}+\mathrm{CB}\) When equilibrium had been established it was found that \(3 / 4\) mole each of reactant \(\mathrm{AB}\) and \(\mathrm{CD}\) has been converted to AD and CB. There is no change in volume. The equilibrium constant of the reaction is? (a) 9 (b) \(\frac{1}{9}\) (c) \(9 / 16\) (d) \(16 / 9\)

The equilibrium constant for the reaction, \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) At temperature \(\mathrm{T}\) is \(4 \times 10^{-4}\). The value of \(\mathrm{K}_{\mathrm{c}}\) for the reaction \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) at the same temperature is: (a) \(4 \times 10^{-6}\) (b) \(2.5 \times 10^{2}\) (c) \(0.02\) (d) 50
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