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What is the \(\mathrm{pH}\) of the solution when \(0.20\) mole of hydrochloric acid is added to one litre of a solution containing. (i) \(1 \mathrm{M}\) each of acetic acid and acetate ion? (ii) \(0.1 \mathrm{M}\) each of acetic acid and acetate ion? Assume the total volume is one litre. \(K_{a}\) for acetic acid \(=1.8 \times 10^{-5}\)

Short Answer

Expert verified
The pH is 4.44 for part (i) and 1 for part (ii) after adding HCl.

Step by step solution

01

Understanding the Problem

We are tasked to calculate the pH of a buffer solution after adding 0.20 moles of HCl to it. The buffer consists of acetic acid (CH鈧僀OOH) and acetate ion (CH鈧僀OO鈦), and we will analyze two different scenarios based on their concentrations.
02

Calculating pH for Part (i) Before Adding HCl

Initially, you have a buffer with 1 M acetic acid and 1 M acetate ion. Use the Henderson-Hasselbalch equation: \[ \mathrm{pH} = \mathrm{p}K_a + \log\left(\frac{[\text{acetate ion}]}{[\text{acetic acid}]}\right) \] Given that the concentrations are equal, \[ \mathrm{pH} = \mathrm{p}K_a = -\log(1.8 \times 10^{-5}) \approx 4.74 \]
03

Applying HCl to the Buffer for Part (i)

HCl provides H鈦 ions that react with the acetate (CH鈧僀OO鈦) to form more acetic acid (CH鈧僀OOH). The change in moles due to 0.20 moles of HCl will be: - Acetate ion decreases to \(1 - 0.20 = 0.80\) mol,- Acetic acid increases to \(1 + 0.20 = 1.20\) mol.
04

Recalculating pH for Part (i) After Adding HCl

Recalculate the pH with the updated concentrations using the Henderson-Hasselbalch equation: \[ \mathrm{pH} = 4.74 + \log\left(\frac{0.80}{1.20}\right) \approx 4.44 \]
05

Calculating pH for Part (ii) Before Adding HCl

Initially, for 0.1 M acetic acid and 0.1 M acetate, the pH is: \[ \mathrm{pH} = 4.74 + \log\left(\frac{0.1}{0.1}\right) = 4.74 \]
06

Applying HCl to the Buffer for Part (ii)

HCl reacts again with the acetate, but since there is 0.1 mole of each component, after adding 0.20 moles of HCl, most acetate reacts and the solution becomes dominated by excess strong acid. Thus: - Acetate ion drops to zero,- Acetic acid becomes \(0.3\) mol (0.1 initial + 0.2 from reaction), and 0.1 mol H鈦 remains.
07

Recalculating pH for Part (ii) After Adding HCl

Since excess H鈦 remains, the pH is governed by these ions: \[ [\text{H}^{+}] = 0.1 \] The pH is \[ \text{pH} = -\log(0.1) = 1 \]
08

Final pH Determinations

For situation (i), after adding HCl, the pH is approximately 4.44 due to the buffer capacity. For situation (ii), the pH drops significantly to 1 due to excess strong acid.

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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 an essential tool in understanding buffer solutions and pH calculations. It provides a simple way to calculate the pH of a solution containing both an acid and its conjugate base. The equation is: \[\mathrm{pH} = \mathrm{p}K_a + \log\left( \frac{[\text{Base}]}{[\text{Acid}]} \right)\]This equation requires the knowledge of the dissociation constant of the acid, known as the acid dissociation constant, \(K_a\), and the concentrations of the acid and its conjugate base. - The term \(\mathrm{p}K_a\) is the negative logarithm of the \(K_a\) value. This provides a measure of the strength of the acid.- The ratio \(\frac{[\text{Base}]}{[\text{Acid}]}\) indicates the relative amounts of the base and acid in the buffer.By manipulating this equation, you can understand how changes in concentration, such as those from adding an acid or base, affect the pH of the solution.
Acid-base reactions
Acid-base reactions are integral to understanding how buffer solutions work. These reactions involve the transfer of protons (H鈦 ions) between acids and bases. In a buffer solution, you typically have a weak acid and its conjugate base, or a weak base and its conjugate acid.- When a strong acid like hydrochloric acid (HCl) is added to a buffer solution, it provides excess H鈦 ions.- These additional H鈦 ions tend to react with the base present in the buffer, in our case, the acetate ion (CH鈧僀OO鈦).The reaction: \[ \text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH} \]This reaction reduces the amount of the conjugate base while increasing the amount of the weak acid, helping to maintain the pH of the solution through these acid-base interactions.
Chemical equilibrium
Chemical equilibrium is a fundamental concept in understanding how buffer solutions resist changes in pH. In a buffer, the acid and base components work together to stabilize the pH through dynamic equilibrium. - Equilibrium involves the forward and reverse reactions of acids and their conjugate bases. - Any addition of a strong acid or base shifts this equilibrium, according to Le Chatelier鈥檚 Principle. In a buffer solution like the one in our exercise: - Initially, you have an equilibrium established between acetic acid (CH鈧僀OOH) and acetate ion (CH鈧僀OO鈦). - When HCl is added, the increase in H鈦 ions causes the equilibrium to shift. The system responds by converting the added H鈦 into more CH鈧僀OOH, balancing the concentrations and resisting drastic pH changes.
pH calculation
Calculating the pH of a solution is crucial when dealing with acid-base reactions and buffer solutions. The pH is a measure of the acidity or basicity of a solution, defined as the negative logarithm of the hydrogen ion concentration:\[ \text{pH} = -\log [\text{H}^+] \]In the exercise:- You first calculate the initial pH using the Henderson-Hasselbalch equation.- After the addition of HCl, you must recalculate pH based on the new concentrations of the acid and conjugate base.For the buffer in part (i) with concentrations of 1 M, the pH changes slightly due to the buffer capacity, shifting from 4.74 to about 4.44.For part (ii) with lower concentrations (0.1 M), the buffer's capacity is exceeded once the H鈦 from HCl is added, resulting in a significant pH drop to 1. This demonstrates how important it is to consider both buffer capacity and the strength of any acids or bases added when calculating pH.

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Most popular questions from this chapter

\(3 \mathrm{~g}\) of acetic acid is added to \(250 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{HCl}\) and the solution made up to \(500 \mathrm{~mL}\). To \(20 \mathrm{~mL}\) of this solution \(\frac{1}{2} \mathrm{~mL}\) of \(5 \mathrm{M} \mathrm{NaOH}\) is added. The \(\mathrm{pH}\) of the solution is [Given: pKa of acetic acid \(=4.75\), molar mass of acetic acid \(=60 \mathrm{~g} / \mathrm{mol}\), $$ \log 3=0.4771] $$ Neglect any changes in volume.

If the equilibrium constant for \(\mathrm{A} \rightleftharpoons \mathrm{B}+\mathrm{C}\) is \(\mathrm{K}_{\mathrm{eq}}^{(1)}\) and that of \(\mathrm{B}+\mathrm{C} \rightleftharpoons \mathrm{P}\) is \(\mathrm{K}_{\mathrm{eq}}^{(2)}\), the equilibrium constant for \(\mathrm{A} \rightleftharpoons \mathrm{P}\) is : (a) \(\mathrm{K}_{\mathrm{eq}}^{(1)} / \mathrm{K}_{\mathrm{eq}}^{(2)}\) (b) \(\mathrm{K}_{\mathrm{eq}}^{(2)}-\mathrm{K}_{\mathrm{eq}}^{(1)}\) (c) \(\mathrm{K}_{\mathrm{eq}}^{(1)}+\mathrm{K}_{\mathrm{eq}}^{(2)}\) (d) \(\mathrm{K}_{\mathrm{eq}}^{(\mathrm{I})} \mathrm{K}_{\mathrm{eq}}^{(2)}\)

An aqueous solution contains \(0.10 \mathrm{MH}_{2} \mathrm{~S}\) and \(0.20 \mathrm{M} \mathrm{HCl}\). If the equilibrium constants for the formation of \(\mathrm{HS}^{-}\)from \(\mathrm{H}_{2} \mathrm{~S}\) is \(1.0 \times 10^{-7}\) and that of \(\mathrm{S}^{2}\) from HS ions is \(1.2 \times 10^{-13}\) then the concentration of \(\mathrm{S}^{2-}\) ions in aqueous solution is : (a) \(5 \times 10^{-8}\) (b) \(3 \times 10^{-20}\) (c) \(6 \times 10^{-21}\) (d) \(5 \times 10^{-19}\)

The set with correct order of acidity is (a) \(\mathrm{HClO}<\mathrm{HClO}_{2}<\mathrm{HClO}_{3}<\mathrm{HClO}_{4}\) (b) \(\mathrm{HClO}_{4}<\mathrm{HClO}_{3}<\mathrm{HClO}_{2}<\mathrm{HClO}\) (c) \(\mathrm{HClO}<\mathrm{HClO}_{4}<\mathrm{HClO}_{3}<\mathrm{HClO}_{2}\) (d) \(\mathrm{HClO}_{4}<\mathrm{HClO}_{2}<\mathrm{HClO}_{3}<\mathrm{HClO}\)

For the reaction, \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) \(\mathrm{DH}=-57.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(\mathrm{K}_{\mathrm{c}}=1.7 \times 10^{16}\) Which of the following statement is INCORRECT? [Main April 10, 2019 (II)] (a) The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required. (b) The equilibrium will shift in forward direction as the pressure increases. (c) The equilibrium constant decreases as the temperature increases. (d) The addition of inert gas at constant volume will not affect the equilibrium constant.

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