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\(500 \mathrm{ml}\) 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. \(\mathrm{K}\) 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 2.0 units.

Step by step solution

01

Calculate Initial pH of the Mixture

Initially, the solution is a mix of HCl and acetic acid, both at 0.2 M. Since HCl is a strong acid, it will fully dissociate, providing 0.2 M of H鈦 ions. The pH of a strong acid can be calculated by pH = -log[H鈦篯. Therefore, the initial pH of the mixture is pH = -log(0.2).
02

Determine Volume of NaOH Required to Neutralize HCl

We have 500 ml (0.5 L) of 0.2 M HCl, giving us 0.1 mol of H鈦 ions. A 0.1 M NaOH solution means each liter contains 0.1 mol of OH鈦. Therefore, 1 liter of 0.1 M NaOH is needed to neutralize 0.1 mol of H鈦, but we only titrate 25 ml (0.025 L) of the mixture. The NaOH needed for this 25 ml is (0.025/0.5) * 0.1 L = 0.005 L (or 5 ml).
03

Calculate pH After HCl is Neutralized

Once HCl is neutralized in the 25 ml portion with 5 ml of NaOH, only acetic acid remains. The remaining acetic acid behaves as a weak acid. Use the formula for pH of a weak acid: \[pH = -\log\sqrt{Ka \times C_o}\]Where \(C_o\) is the concentration of acetic acid in the portion being titrated. After neutralizing HCl, consider only acetic acid to calculate the change in pH.
04

Determine the Total pH Change

Subtract the pH at the end of neutralization from the initial pH of the mixture to find the change in pH. Using the calculations, this change is approximately 2.0.

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Key Concepts

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

Initial pH Calculation
When two acids are mixed, like in this exercise, it's important first to focus on the stronger acid, which here is hydrochloric acid (HCl). In a mixture with both HCl and acetic acid (CH_3COOH), the HCl will completely ionize into H鈦 ions because it's a strong acid. This complete ionization means all HCl present will contribute to the concentration of hydrogen ions in the solution.
To calculate the initial pH of the solution, we use the formula for pH of a strong acid: \(pH = -\log [H^+]\). Here, [H^+] is given by the molarity of HCl, which is 0.2 M due to complete dissociation. Therefore, substitute [H^+] with 0.2 in the formula to find the initial pH:\(-\log (0.2)\).
This initially calculated pH sets the baseline for understanding how subsequent reactions, such as neutralization, will impact the pH of the solution.
Neutralization Reaction
A neutralization reaction occurs when an acid reacts with a base to form water and a salt. In this exercise, sodium hydroxide (NaOH) is used as a base to neutralize the hydrochloric acid in the solution.
The main concept here involves the stoichiometry of reaction: every mole of HCl reacts with exactly one mole of NaOH. In simpler terms, the H鈦 ions from HCl combine with the OH鈦 ions from NaOH to create water (H鈧侽), removing free hydrogen ions from the solution and influencing the pH.
The volume and molarity of each solution are crucial. We need enough NaOH to match the moles of H鈦 ions present in the 25 ml of the mixture being titrated. Calculating this properly is crucial for ensuring complete neutralization of HCl and assessing how the pH changes as a result.
pH of Weak Acid
After the complete neutralization of HCl, only the weak acid CH_3COOH is left in the solution. Unlike strong acids, weak acids do not fully dissociate in water, which affects the pH calculation.
To find the pH of a weak acid like acetic acid, we use the formula:\[pH = -\log \sqrt{K_a \times C_o}\], where \(K_a\) is the acid dissociation constant and \(C_o\) is the initial concentration. This equation helps estimate how much of the weak acid dissociates into hydrogen ions in the solution.
This consideration is important because as CH_3COOH only partially dissociates, the effect on pH is more gradual, leading to subtler changes than those seen with strong acids.
Acid Dissociation Constant (K_a)
The acid dissociation constant, \(K_a\), signifies the degree to which an acid dissociates in solution. For weak acids, knowing \(K_a\) is fundamental because it enables us to determine the extent of dissociation and helps in calculating the pH.
For acetic acid in our exercise, the \(K_a\) value is provided as \(2.0 \times 10^{-5}\). This low \(K_a\) suggests that acetic acid ionizes only slightly, yielding fewer hydrogen ions than equivalent concentrations of stronger acids would.
By incorporating \(K_a\) into the pH equation for weak acids, students can appreciate the detailed dynamic between acidic strength, ionization, and the resulting impact on pH. Understanding \(K_a\) is critical for chemistry students to grasping acid strength and comparing different acidic reactions strategically.

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Most popular questions from this chapter

Which of the following solutions will have no effect on pH on dilution? (a) \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONa}\) (b) \(1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONH}_{4}\) (c) \(0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}+0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) (d) \(0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{CO}_{3}+0.5 \mathrm{M} \mathrm{NaHCO}_{3}\)

When rain is accompanied by a thunderstorm, the collected rain water will have a \(\mathrm{pH}\) value (a) slightly lower than that of rain water without thunderstorm (b) slightly higher than that when the thunderstorm in not there(c) uninfluenced by occurrence of thunderstorm (d) which depends on the amount of dust in air

Equal volumes of \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) (c M) solution of \(\mathrm{pH}=5\) is mixed with HCl solution of same \(\mathrm{pH}\). Which of the following is an incorrect statement? (a) Concentration of \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) will become \(\mathrm{c} / 2 \mathrm{M}\) after mixing \(\mathrm{HCl}\) with it. (b) Concentration of \(\mathrm{H}^{+}\)after mixing the two solutions is \(10^{-5} \mathrm{M}\). (c) The degree of dissociation of \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) is suppressed due to addition of \(\mathrm{HCl}\). (d) Original concentration of \(\mathrm{HCl}\) was \(10^{-5} \mathrm{M}\). Passage-2 Solubility product of an electrolyte at a particular temperature is defined as the product of conc. of its ions in a saturated solution, each conc. raised to the power equal to the number of ions produced on dissociation of one molecule of the electrolyte. \(\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{y}} \rightleftharpoons \mathrm{xA}^{+}+\mathrm{yB}^{-}\) \(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{+}\right]^{\mathrm{x}}\left[\mathrm{B}^{-}\right]^{\mathrm{y}}\) Ionic product of the electrolyte \(\mathrm{A}_{x} \mathrm{~B}_{y}\) is also equal to \(\left[\mathrm{A}^{+}\right]^{\times}\left[\mathrm{B}^{-}\right]^{\mathrm{y}}\) but it is applicable to all types of solutions, which may be saturated or unsaturated.

The dissociation constant of a substituted benzoic acid is \(1.0 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). The \(\mathrm{pH}\) of \(0.01 \mathrm{~m}\) solution of its sodium salt is

If \(50 \mathrm{ml}\) of \(0.2 \mathrm{M} \mathrm{NaCN}\) is mixed with \(50 \mathrm{ml}\) of \(0.2\) M \(\mathrm{HCl}\), then \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{CN}^{-}\right]=\mathrm{x} \times 10^{-6}\) where \(\mathrm{x}\) is \(\left(\mathrm{K}_{\mathrm{b}}\right.\) for \(\left.\mathrm{CN}^{-}=2 \times 10^{-5}\right)\)

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