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Chapter 8: Problem 153

\(500 \mathrm{ml}\) of \(0.2 \mathrm{M} \mathrm{HCl}\) is mixed with \(500 \mathrm{ml}\) of \(0.2 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH} .25 \mathrm{ml}\) of the mixture is titrated with \(0.1\) M NaOH solution. By how many units does the \(\mathrm{pH}\) change from the start to the stage when \(\mathrm{HCl}\) is just completely neutralized. \(\mathrm{K}_{\mathrm{a}}\) for acetic acid \(=2.0 \times 10^{-5}\). (a) \(3.7\) (b) \(4.4\) (c) \(2.0\) (d) \(3.0\)

Short Answer

Expert verified
The pH change is closest to 2.0 units (option c).

Step by step solution

01

Understanding Initial Conditions

We start with 500 ml of 0.2 M HCl and 500 ml of 0.2 M CH3COOH, so the total volume of the solution is 1000 ml. The concentration of HCl and CH3COOH remain 0.2 M since they are mixed in equal volumes of the same concentration.
02

Calculate Initial Moles of HCl and CH3COOH

The initial moles of HCl are given by multiplying volume by concentration: \( moles = 0.5 imes 0.2 = 0.1 \ mol \). The initial moles for CH3COOH are the same due to equal volume and concentration.
03

Initial pH Calculation

HCl is a strong acid and will dissociate completely. The concentration of H+ will be 0.1 M. Therefore, the initial pH is \( \text{pH} = -\log[H^+] = -\log(0.1) = 1.0 \).
04

Determine NaOH Volume for Neutralization

To neutralize the HCl, the same number of moles of NaOH is needed (0.1 mol). Since 0.1 M NaOH is used, the volume needed is \( \frac{0.1}{0.1} = 1 \) L.
05

Identify Change when HCl is Neutralized

When all HCl is neutralized and the addition of NaOH stops, no HCl contributes to the pH. The remaining acidic property primarily comes from CH3COOH. Now 0.1 M Acetic acid needs to be evaluated.
06

Calculating the New pH

Now only CH3COOH is contributing to the pH. The concentration of CH3COOH is still 0.1 M, thus replace into the equation for weak acids: \[ \text{pH} = \frac{1}{2}(\text{pK}_a - \log[C]) = \frac{1}{2}( -\log(2.0 \times 10^{-5}) - \log 0.1) \]\[ = \frac{1}{2}(4.7 + 1) = \frac{1}{2}(5.7) = 2.85 \]
07

Calculate the pH Change

The difference in pH from the initial state to when HCl is neutralized is the initial pH 1.0 to the calculated pH of 2.85. The difference is 2.85 - 1.0 = 1.85, which is closest to option (c).

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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 method used to determine the concentration of an unknown substance by reacting it with a substance of known concentration. In this exercise, titration involves adding a solution of sodium hydroxide (NaOH) to a mixture of two acids: hydrochloric acid (HCl) and acetic acid (CH鈧僀OOH).
During titration, the NaOH gradually neutralizes the acid by reacting with the hydrogen ions ( 1+ 1 3 5) in the solution, reducing their concentration. The aim is to reach a point where the number of moles of NaOH equals the number of moles of HCl, known as the equivalence point. This process allows us to explore how the pH of the solution changes throughout the reaction.
Neutralization Reaction
A neutralization reaction occurs when an acid and a base react to form water and a salt. In the case of this exercise, the reaction involves HCl, a strong acid, and NaOH, a strong base. The balanced chemical equation representing this reaction is:
  • HCl + NaOH 鈫� NaCl + H鈧侽
This reaction is a classic example of neutralization as the hydronium ions ( 1+ ) from the acid react with hydroxide ions ( 1- ) from the base to form water. As the strong acid HCl completely dissociates, its pH can be straightforwardly calculated at the start of the reaction. Neutralization is complete when all the HCl is consumed by the NaOH, and only the weak acid CH鈧僀OOH remains affecting the solution's pH.
Weak Acid Dissociation
Weak acids like acetic acid (CH鈧僀OOH) do not fully dissociate in water. Instead, they establish an equilibrium between the undissociated acid and its ions:
  • CH鈧僀OOH 鈬� CH鈧僀OO鈦� + H鈦�
Unlike strong acids, calculating the pH of a weak acid requires using its dissociation constant ( 1+ ]). This constant indicates the strength of the acid and how well it can donate H鈦� ions. For acetic acid, the dissociation constant, A, is quite low at 2.0 x 10 5, indicating it only partially dissociates.
This means even after the complete neutralization of HCl via titration, the acetic acid still contributes to the pH of the solution, as only a portion of it releases hydrogen ions.
HCl and CH鈧僀OOH Mixture
In this exercise, a mixture of hydrochloric acid and acetic acid is used, which alters the solution's behavior in titration. Initially, both acids contribute to the pH. However, since HCl is a strong acid, it dissociates completely, taking precedence over the weakly dissociating acetic acid in the initial pH calculation.
Once the HCl is neutralized by NaOH, the CH鈧僀OOH becomes the primary contributor to acidity due to its still-present undissociated form. This change in contribution post-neutralization showcases how mixtures of strong and weak acids can complicate pH calculations, even when initially they appear straightforward.
pKa and pH Relationship
The pH of a solution and the pKa of an acid are closely related. The pKa is the negative logarithm of the acid dissociation constant, A, which helps in understanding the acid's strength. For weak acids like acetic acid, the pKa is frequently used to calculate pH.
The formula for determining the pH of a weak acid like CH鈧僀OOH is:
  • pH = (pKa + og C
This structure highlights the relationship between pH, the concentration of the acid ( A-CH鈧僀OOH 6), and its pKa. Understanding this balance is key to determining the impact of weak acids on the overall pH, particularly in complex solutions like mixtures. In this exercise, it elucidates why the change in pH after neutralizing HCl was calculated using the pKa of acetic acid, showcasing its continued influence on pH.

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Most popular questions from this chapter

The pKa of acetyl salicylic acid (aspirin) is \(3.5\). The \(\mathrm{pH}\) of gastric juice in human stomach is about \(2-3\) and \(\mathrm{pH}\) in the small intestine is about 8 . Aspirin will be (a) ionized in the small intestine and almost unionized in the stomach (b) unionized in the small intestine and in the stomach (c) completely ionized in the small intestine and in the stomach (d) ionized in the stomach and almost unionized in the small intestine.

The \(\mathrm{pH}\) of \(\mathrm{N} / 10 \mathrm{HCl}\) is (a) 3 (b) 4 (c) 2 (d) 1

At \(90^{\circ} \mathrm{C}\), pure water has \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] 10^{-6}\) mole litre \(^{-1} .\) What is the value of \(\mathrm{K}_{\mathrm{w}}\) at \(90^{\circ} \mathrm{C}\) ? (a) \(10^{-8}\) (b) \(10^{-6}\) (c) \(10^{-12}\) (d) \(10^{-14}\)

The precipitate of \(\mathrm{CaF}_{2}\left(\mathrm{Ksp}=1.7 \times 10^{-10}\right)\) is obtained when equal volumes of the following are mixed (a) \(10^{-2} \mathrm{M} \mathrm{Ca}^{2+}+10^{-3} \mathrm{MF}^{-}\) (b) \(10^{-3} \mathrm{M} \mathrm{Ca}^{2+}+10^{-5} \mathrm{MF}^{-}\) (c) \(10^{-4} \mathrm{M} \mathrm{Ca}^{2+}+10^{-4} \mathrm{MF}^{-}\) (d) \(10^{-5} \mathrm{M} \mathrm{Ca}^{2+}+10^{-3} \mathrm{MF}^{-}\)

A buffer solution can be prepared from a mixture of 1\. sodium acetate and acetic acid in water 2\. sodium acetate and hydrochloric acid in water 3\. ammonia and ammonium chloride in water 4\. ammonia and sodium hydroxide in water. (a) \(1,3,4\) (b) \(2,3,4\) (c) \(1,2,4\) (d) \(1,2,3\)
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