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Chapter 7: Problem 41

A certain buffer solution contains equal concentration of \(X^{-}\)and \(H X\). The \(K_{b}\) for \(X^{-}\)is \(10^{-10} .\) The \(\mathrm{pH}\) of the buffer is: (a) 4 (b) 7 (c) 10 (d) 14

Short Answer

Expert verified
The pH of the buffer is 4.

Step by step solution

01

Understand the Problem

The problem is asking for the pH of a buffer solution that contains equal concentrations of the base form \(X^{-}\) and its conjugate acid \(HX\). We are given the base dissociation constant \(K_b\) for \(X^{-}\).
02

Identify Key Formulas

In a buffer solution where the concentrations of the conjugate acid-base pair are equal, the pH can be related to the \(pK_a\) of the acid. The formula is derived from the Henderson-Hasselbalch equation. We need to convert \(K_b\) to \(K_a\) to find the \(pH\).
03

Calculate \(K_a\) from \(K_b\)

Use the relation \(K_w = K_a \times K_b\) where \(K_w = 10^{-14}\) at 25掳C. Rearrange to solve for \(K_a\): \[ K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{10^{-10}} = 10^{-4} \]
04

Determine \(pK_a\)

Calculate \(pK_a\) using the formula \(pK_a = -\log K_a\):\[ pK_a = -\log(10^{-4}) = 4 \]
05

Conclude the pH

Since the concentrations of \(X^{-}\) and \(HX\) are equal, the pH of the buffer solution is equal to \(pK_a\). Therefore, the pH = 4.

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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 used in chemistry to estimate the pH of a buffer solution. A buffer solution is capable of resisting changes in pH, even when an acid or base is added. This equation is a significant tool that simplifies the calculations needed to find the pH of solutions containing a weak acid and its conjugate base, or a weak base and its conjugate acid.
  • The equation is written as: \[ \text{pH} = \text{p}K_a + \log \left( \frac{[A^-]}{[HA]} \right) \] where \([A^-]\) represents the concentration of the base form and \([HA]\) the concentration of the acid form.
  • For situations where the concentrations of the acid and base forms are equal, the log ratio becomes zero, and thus, \(\text{pH} = \text{p}K_a\).
This makes the equation particularly useful in buffer solutions because you can directly relate the pH to the pKa of the weak acid in the solution.
pH calculation
Calculating pH is an essential part of chemistry, especially when dealing with acidic or basic solutions. The pH is a scale used to specify the acidity or basicity of an aqueous solution. It is the negative logarithm (base 10) of the activity of hydrogen ions (\(H^+\)) in a solution.
  • In practical terms, pH = -log[\(H^+]\). A pH less than 7 indicates an acidic solution, whereas a pH greater than 7 indicates a basic solution.
  • For neutral solutions, like pure water at 25掳C, the pH is exactly 7.
When utilizing buffer solutions, the pH can be calculated directly using the Henderson-Hasselbalch equation, if the solution's composition is known. This is especially helpful because it allows chemists to predict and maintain conditions suitable for various biological or chemical processes.
Base dissociation constant (K_b)
The base dissociation constant, \(K_b\), is a measure of the strength of a base in solution. It indicates the degree to which a base can dissociate into its ions in water, with larger values representing stronger bases.
  • \(K_b\) is a critical factor when determining the pH of solutions containing bases. It reflects the base's ability to accept protons from water, forming its conjugate acid and hydroxide ions.
  • The general expression for the base dissociation constant is: \[ K_b = \frac{[B^+][OH^-]}{[BOH]} \] where \([B^+]\) is the concentration of the conjugate acid, \([OH^-]\) is the concentration of hydroxide ions, and \([BOH]\) is the concentration of the base itself.
For buffers involving weak bases, \(K_b\) helps in determining the corresponding \(K_a\), which in turn assists in finding the pH of the buffer solution.
Acid dissociation constant (K_a)
The acid dissociation constant, \(K_a\), quantifies the strength of an acid in terms of its ability to donate protons to the solvent, typically water. Similar to \(K_b\), a higher \(K_a\) value means a stronger acid.
  • For any given acidic reaction where the acid (HA) dissociates into \(H^+\) (hydrogen ion) and \(A^-\) (anion), the expression is: \[ K_a = \frac{[H^+][A^-]}{[HA]} \]
  • \(K_a\) is connected to \(K_b\) through the water dissociation constant ( \(K_w = 10^{-14}\) at 25掳C), such that \(K_w = K_a \times K_b\). This relationship is pivotal for determining the missing dissociation constant when only one is known.
In buffer solutions, knowing \(K_a\) is crucial for using the Henderson-Hasselbalch equation to calculate the pH accurately. This relationship ensures a comprehensive understanding of the solution's behavior.

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Most popular questions from this chapter

The \(\mathrm{pH}\) of \(0.1 \mathrm{M}\) solution of the following salts increases in the order. (a) \(\mathrm{NaCl}<\mathrm{NH}_{4} \mathrm{Cl}<\mathrm{NaCN}<\mathrm{HCl}\) (b) \(\mathrm{HCl}<\mathrm{NH}_{4} \mathrm{Cl}<\mathrm{NaCl}<\mathrm{NaCN}\) (c) \(\mathrm{NaCN}<\mathrm{NH}_{4} \mathrm{Cl}<\mathrm{NaCl}<\mathrm{HCl}\) (d) \(\mathrm{HCl}<\mathrm{NaCl}<\mathrm{NaCN}<\mathrm{NH}_{4} \mathrm{Cl}\)

The Haber's process for the formation of \(\mathrm{NH}_{3}\) at \(298 \mathrm{~K}\) is \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} ; \Delta H=-46.0 \mathrm{~kJ} ;\) Which of the following is the correct statement (a) The condition for equilibrium is $$ G_{\mathrm{N}_{2}}+3 G_{\mathrm{H}_{2}}=2 G_{\mathrm{NH}_{3}} $$ where \(G\) is Gibb's free energy per mole of the gaseous species measured at that partial pressure. (b) On adding \(\mathrm{N}_{2}\), the equilibrium will shift to forward direction because according to \(\mathrm{II}^{\text {nd }}\) law of thermodynamics, the entropy must increase in the direction of spontaneous reaction (c) The catalyst will increase the rate of forward reaction by 2 times and that of backward reaction by \(1.5\) times (d) None of these

Amongst the following hydroxides, the one which has the lowest value of \(K_{s p}\) at ordinary temperature (about \(25^{\circ} \mathrm{C}\) ) is [1990-1 Mark] (a) \(\mathrm{Mg}(\mathrm{OH})_{2}\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}\) (c) \(\mathrm{Ba}(\mathrm{OH})_{2}\) (d) \(\mathrm{Be}(\mathrm{OH})_{2}\)

The value of \(\mathrm{Kc}\) is 64 at \(800 \mathrm{~K}\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) .\) The value of \(K_{\mathrm{c}}\) for the following reaction is : \(\mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g})\) (a) \(1 / 64\) (b) 8 (c) \(1 / 4\) (d) \(1 / 8\)

The compound that is not a Lewis acid is : (a) \(\mathrm{BF}_{3}\) (b) \(\mathrm{AlCl}_{3}\) (c) \(\mathrm{BeCl}_{2}\) (d) \(\mathrm{SnCl}_{4}\)
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