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Chapter 8: Problem 172

Calculate the ratio of \(\mathrm{pH}\) of a solution containing 1 mole of \(\mathrm{CH}_{3} \mathrm{COONa}+1\) mole of \(\mathrm{HCl}\) per litre to that of a solution containing 1 mole of \(\mathrm{CH}_{3} \mathrm{COONa}+1\) mole of \(\mathrm{CH}_{3} \mathrm{COOH}\) per litre. (a) \(\frac{2}{1}\) (b) \(\frac{1}{2}\) (c) \(\frac{2}{3}\) (d) \(\frac{3}{2}\)

Short Answer

Expert verified
The correct answer is (c) \(\frac{2}{3}\).

Step by step solution

01

Identify Reactants in Solution 1

In the first solution, we have 1 mole of CH鈧僀OONa and 1 mole of HCl per litre. CH鈧僀OONa dissociates into CH鈧僀OO鈦� and Na鈦� ions. HCl completely dissociates into H鈦� and Cl鈦� ions.
02

Determine Reaction in Solution 1

H鈦� ions from HCl will react with CH鈧僀OO鈦� ions to form CH鈧僀OOH:\[\text{CH鈧僀OO鈦粆 + \text{H鈦簘 \rightarrow \text{CH鈧僀OOH}\]Since both H鈦� and CH鈧僀OO鈦� are equimolar, they completely react, resulting in a 1 molar solution of CH鈧僀OOH.
03

Calculate pH of Solution 1

Solution 1 is now a pure 1 molar solution of CH鈧僀OOH, a weak acid. The pH can be obtained from its dissociation constant, but since the whole concentration is converted to acid, we approximate it as pH = -log [H鈦篯. For typical acetic acid, this simplifies to around pH = 2 assuming full conversion.
04

Identify Composition in Solution 2

In the second solution, we have 1 mole of CH鈧僀OONa and 1 mole of CH鈧僀OOH per litre. These two form a buffer solution, where the concentrations of the acetate ion and acetic acid remain in a 1:1 ratio.
05

Calculate pH of Solution 2 using Henderson-Hasselbalch Equation

For a buffer solution of acetic acid and sodium acetate:\[\text{pH} = \text{pK}_a + \log \left(\frac{[\text{Base}]}{[\text{Acid}]}\right)\]Since the base (CH鈧僀OO鈦� from CH鈧僀OONa) and the acid (CH鈧僀OOH) have the same concentration, the ratio is 1 and \(\log(1) = 0\). Therefore, \(\text{pH} = \text{pK}_a\) of CH鈧僀OOH, which is around 4.76.
06

Calculate the Ratio of pH Values

The pH of solution 1 is about 2 and the pH of solution 2 is about 4.76. The ratio of their pH values is:\[\frac{2}{4.76} \approx \frac{2}{3}\]Thus, the ratio is closest to \(\frac{2}{3}\).

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Key Concepts

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

pH Calculation
Calculating the pH of a solution is a fundamental concept in chemistry, as it gives us an idea of how acidic or basic the solution is. The "pH" stands for "potential of hydrogen" and is a measure of the concentration of hydrogen ions (\(\text{H}^+\)) in a solution. The formula for calculating pH is:
\[ \text{pH} = -\log[\text{H}^+] \]where [H鈦篯 is the molarity of hydrogen ions.

For example, if the concentration of hydrogen ions in a solution is 0.01 M, the pH would be calculated as:
\[ \text{pH} = -\log(0.01) = 2 \]
In this exercise, we calculated the pH of two solutions. The first solution was identified as having a lower pH because it became a strong acid solution, and the second solution had a higher pH as it formed a buffer. These calculations help in understanding the comparative acidity and basicity of different solutions.
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch Equation is an important tool for calculating the pH of buffer solutions. A buffer solution can resist changes in pH when acids or bases are added to it. It usually consists of a weak acid and its conjugate base.

The equation is expressed as:
\[\text{pH} = \text{pK}_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right)\]
where:
  • \( \text{pK}_a \) is the negative logarithm of the acid dissociation constant, \( K_a \), of the weak acid.
  • [Base] is the concentration of the conjugate base.
  • [Acid] is the concentration of the weak acid.
In our exercise, the second solution formed a buffer with equal concentrations of acetic acid (CH鈧僀OOH) and acetate ions (CH鈧僀OO鈦�), which simplified the expression because the log of 1 is 0. Thus, the pH remained equal to the \( \text{pK}_a \) of acetic acid, which is approximately 4.76.
Acid-Base Reactions
Acid-base reactions are essential in understanding how different chemical species interact in solution. A typical acid-base reaction involves the transfer of protons (\(\text{H}^+\)) from an acid to a base.

For instance, when hydrochloric acid (HCl), a strong acid, reacts with sodium acetate (CH鈧僀OONa), a source of acetate ions, the \(\text{H}^+\) ions from HCl will combine with acetate ions (CH鈧僀OO鈦�) to form acetic acid (CH鈧僀OOH):
\[\text{CH鈧僀OO}^- + \text{H}^+ \rightarrow \text{CH鈧僀OOH}\]
In this way, the strong acid completely neutralizes the base, converting it into a weak acid.

This is what occurred in the first solution of our exercise. In contrast, the second solution's acid and base were already in equimolar amounts, serving as a stabilizing buffer without any significant net reaction, which maintained its pH more stable. Understanding these reactions enables chemists to predict and control the outcomes in diverse chemical environments.

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Most popular questions from this chapter

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 sparingly soluble salt \(\mathrm{M}(\mathrm{OH})_{\mathrm{x}}\) has \(\mathrm{K}_{\mathrm{sp}}=4 \times 10^{-12}\). Its solubility is \(10^{-4} \mathrm{M}\). The value of \(\mathrm{x}\) is

When \(60 \mathrm{ml}\) of \(0.1 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) is mixed with \(40 \mathrm{ml}\) of \(0.125 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}, \mathrm{CaCO}_{3}\) precipitates. If \(\mathrm{K}_{s p}\) of \(\mathrm{CaCO}_{3}\) is \(5 \times 10^{-9} \mathrm{M}^{2}\), the \(\left[\mathrm{CO}_{3}^{2-}\right]\) in the resulting solution is (a) \(5 \times 10^{8} \mathrm{M}\) (b) \(5 \times 10^{-9} \mathrm{M}\) (c) \(5 \times 10^{-6} \mathrm{M}\) (d) \(5 \times 10^{-7} \mathrm{M}\)

For the reaction \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-(\mathrm{aq})} \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+2 \mathrm{CN}^{-}\) (aq), the equilibrium constant at \(25^{\circ} \mathrm{C}\) is \(4.0 \times 10^{-19}\). Calculate the silver ion concentration in a solution which was originally \(0.10 \mathrm{M}\) in \(\mathrm{KCN}\) and \(0.03 \mathrm{M}\) in \(\mathrm{AgNO}_{3}\). (a) \(2.5 \times 10^{-18} \mathrm{M}\) (b) \(1.5 \times 10^{-18} \mathrm{M}\) (c) \(5.5 \times 10^{-18} \mathrm{M}\) (d) \(7.5 \times 10^{-18} \mathrm{M}\)

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) \(0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{CO}_{3}+0.5 \mathrm{M} \mathrm{NaHCO}_{3}\)
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