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Chapter 7: Problem 20

Which one of the following statement is correct? (a) Bronsted-Lowry theory could not explain the acidic nature of \(\mathrm{BCl}_{3}\) (b) The \(\mathrm{pH}\) of \(0.01 \mathrm{M} \mathrm{NaOH}\) solution is 2 (c) The ionic product of water at \(25^{\circ} \mathrm{C}\) is \(10^{-10} \mathrm{~mol}^{2} \mathrm{~L}^{-2}\) (d) The \(\mathrm{pH}\) of a solution can be calculated using the equation \(\mathrm{pH}=\log \left[\mathrm{H}^{+}\right]\)

Short Answer

Expert verified
Option (a) is correct; BCl3's acidic nature is not explained by Bronsted-Lowry theory.

Step by step solution

01

Analyze Option (a)

Bronsted-Lowry theory defines acids as proton donors. BCl3 (boron trichloride) does not donate protons in solution; it is a Lewis acid because it can accept electron pairs. Therefore, this statement is correct as Bronsted-Lowry theory does not explain the acidity of BCl3.
02

Evaluate Option (b)

The pH of a solution is calculated using the formula \( \text{pH} = -\log[\text{H}^+] \). For 0.01 M NaOH, the OH ion concentration is 0.01 M, so \( [\text{H}^+] = \frac{10^{-14}}{0.01} = 10^{-12} \). The pH would be 12, not 2. Thus, this statement is incorrect.
03

Verify Option (c)

At 251C, the ionic product of water \( K_w \) is \( 1.0 \times 10^{-14} \text{ mol}^2 \text{ L}^{-2} \). The given value of \( 10^{-10} \text{ mol}^2 \text{ L}^{-2} \) is incorrect.
04

Check Option (d)

The pH is calculated based on the concentration of hydrogen ions using the formula \( \text{pH} = -\log[\text{H}^+] \). The given equation \( \text{pH} = \log[\text{H}^+] \) is incorrect because it misses the minus sign.

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Key Concepts

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

Bronsted-Lowry theory
The Bronsted-Lowry theory is an essential concept in understanding acids and bases. According to this theory:
  • An acid is a substance that donates a proton ( H鈦� ) to another substance.
  • A base is a substance that receives a proton ( H鈦� ) from an acid.
Boron trichloride ( BCl鈧� ) presents an interesting case when evaluated through this theory. Since BCl鈧� doesn't donate a proton, it doesn't fit into the Bronsted-Lowry definition of an acid. Instead, BCl鈧� acts as a Lewis acid, which is defined as a compound that can accept an electron pair. So, while Bronsted-Lowry theory is very useful, it doesn't cover all acidic substances, especially those like BCl鈧� , which require the Lewis theory for explanation.
Lewis acids
Lewis acids and bases introduce a broader scope of what can be considered acidic or basic. A Lewis acid is an electron pair acceptor, meaning it accepts a pair of electrons during a chemical reaction. In contrast, a Lewis base donates an electron pair.
  • BCl鈧� , for instance, functions as a Lewis acid because it can accept an electron pair due to its empty p-orbital.
  • Lewis theory expands the definition of acids beyond just proton donation.
  • This is particularly useful when analyzing complex molecules that do not fit the conventional acid descriptions.
Lewis acids play crucial roles in many chemical reactions, including catalysis and organic synthesis, making this theory fundamental in many advanced fields of chemistry beyond simple ionic solutions.
pH calculations
The concept of pH is a way to measure the acidity or basicity of a solution. The pH scale ranges from 0 to 14. Solutions with a pH less than 7 are acidic, and those with a pH greater than 7 are basic.

The formula to calculate pH is: \( ext{pH} = - ext{log}[ ext{H}^+] \). This means that pH is the negative logarithm of the hydrogen ion concentration in a solution.
  • For a 0.01 ext{M} NaOH solution, the concentration of hydroxide ions [OH^-] is 0.01 ext{M}.
  • The concentration of hydrogen ions [H^+] can be calculated through the relation [K_w = [H^+][OH^-], where K_w=1.0 imes 10^{-14} ext{ mol}^2 ext{ L}^{-2} at 25^{ ext{鈭榼}C. Thus, [H^+]= rac{1.0 imes 10^{-14}}{0.01}.
  • The pH calculates to 12, confirming that the solution is indeed basic.
The minus sign in the formula is crucial as it defines the very nature of the log scale employed in measuring acidity.
Ionic product of water
In chemistry, the ionic product of water, denoted as K_w , is critical for understanding the nature of aqueous solutions. It quantifies how water ionizes into hydrogen ions and hydroxide ions:
  • The equation is K_w=[H^+][OH^-] .
  • At 25^{ ext{鈭榼} C, K_w typically equals 1.0 imes 10^{-14} ext{ mol}^2 ext{ L}^{-2} .
  • This means that in pure water, the concentrations of [H^+] and [OH^-] are each 1.0 imes 10^{-7} ext{ M} .
This constant explains why pure water has a pH of 7, which is considered neutral. The K_w value changes with temperature, affecting the equilibrium position of water ionization. Understanding K_w allows chemists to calculate the pH of various solutions and assess the effect of temperature on ionic equilibrium.

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Most popular questions from this chapter

In which of the following cases does the reaction go farthest to completion: (a) \(\mathrm{K}=1\) (b) \(\mathrm{K}=10\) (c) \(\mathrm{K}=10^{-2}\) (d) \(\mathrm{K}=10^{2}\)

If equilibrium constant for the reaction, \(\mathrm{XO}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1) \rightleftharpoons \mathrm{HXO}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\) is \(0.36 \times 10^{-6}\) then find the value of dissociation constant \(\left(\mathrm{K}_{\mathrm{a}}\right)\) for \(\mathrm{HXO}:\) (a) \(0,36 \times 10^{-8}\) (b) \(2.8 \times 10^{-8}\) (c) \(2.8 \times 10^{-10}\) (d) \(0.36 \times 10^{-6}\)

Which of the following reaction will be favoured at low pressure: (a) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) (b) \(\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\) (c) \(\mathrm{PCl}_{5} \rightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}\) (d) \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\)

The rate constants for the forward and backward reactions of hydrolysis of ester are \(1.1 \times 10^{-2}\) and \(1.5 \times\) \(10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) respectively. The equilibrium constant of the reaction, \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}+\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is: (a) \(6.53\) (b) \(7.34\) (c) \(7.75\) (d) \(8.33\)

If equilibrium constants of reaction: \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\) is \(\mathrm{K}_{1}\), and \(\frac{1}{2} \mathrm{~N}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{NO}\) is \(\mathrm{K}_{2}\) then (a) \(\mathrm{K}_{1}=\mathrm{K}_{2}\) (b) \(\mathrm{K}_{1}=2 \mathrm{~K}_{2}\) (c) \(\mathrm{K}_{2}=\sqrt{\mathrm{K}}_{1}\) (d) \(\mathrm{K}_{1}=\frac{1}{2} \mathrm{~K}_{2}\)
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