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Chapter 5: Problem 128

What quantity of sodium acetate must be added to \(1.00\) litre of a \(0.200\) ?M acetic acid \(\left(\mathrm{K}_{u}=1.8 \times 10^{-5}\right)\) solution to form a buffer of \(\mathrm{pH}\) \(=4.30 ?\) a. \(7.18 \times 10^{-2}\) moles b. \(3.62 \times 10^{-5}\) moles c. \(7.18 \times \mathrm{v} 10^{-3}\) moles d. \(4.15 \times 10^{-9}\) moles

Short Answer

Expert verified
Add 7.18 脳 10^{-2} moles of sodium acetate.

Step by step solution

01

Understand the Buffer Equation

To calculate the quantity of sodium acetate needed, we'll apply the Henderson-Hasselbalch equation for buffers, given by:\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]where \([\text{A}^-]\) is the concentration of the acetate ion (from sodium acetate), \([\text{HA}]\) is the concentration of the acetic acid, and \(\text{pKa} = -\log(\text{K}_a)\).
02

Calculate the pKa of Acetic Acid

First, calculate the pKa of acetic acid:\[pKa = -\log(1.8 \times 10^{-5}) \]Using a calculator, we find that:\[ pKa \approx 4.74 \]
03

Set Up the Henderson-Hasselbalch Equation

With \(\text{pH} = 4.30\) and \(\text{pKa} = 4.74\), plug these values into the Henderson-Hasselbalch equation:\[4.30 = 4.74 + \log \left( \frac{[\text{A}^-]}{0.200} \right) \]
04

Solve for the Ratio [A^-]/[HA]

Rearrange the equation to solve for the ratio of conjugate base to acid:\[\log \left( \frac{[\text{A}^-]}{0.200} \right) = 4.30 - 4.74 = -0.44 \]Convert the logarithmic equation to its exponential form:\[\frac{[\text{A}^-]}{0.200} = 10^{-0.44} \]Calculate this ratio using a calculator:\[\frac{[\text{A}^-]}{0.200} \approx 0.363 \]
05

Calculate the Concentration of Acetate Ion [A^-]

Multiply both sides by \(0.200\) to solve for \([\text{A}^-]\):\[[\text{A}^-] = 0.363 \times 0.200 \]Calculate to find:\[[\text{A}^-] \approx 0.0726 \, \text{M} \]
06

Calculate the Moles of Sodium Acetate Needed

Since the volume of the solution is \(1.00\) liter, the moles of sodium acetate needed is simply the concentration found above:\[\text{moles of acetate} = 0.0726 \, \text{M} \times 1.00 \, \text{L} = 0.0726 \, \text{moles} \]Rounding to match the options, we find:\[ \text{moles of acetate} \approx 7.18 \times 10^{-2} \, \text{moles} \]
07

Identify the Correct Answer Choice

Compare this calculated value to the given choices:a. \(7.18 \times 10^{-2}\) molesb. \(3.62 \times 10^{-5}\) molesc. \(7.18 \times 10^{-3}\) molesd. \(4.15 \times 10^{-9}\) molesThe correct answer is choice **a.**

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Key Concepts

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

Buffers
Buffers are like cushions for pH fluctuations in solutions. They maintain a stable pH when small amounts of acids or bases are added. You can think of buffers as protective shields. In a buffer, there is usually a weak acid and its conjugate base.
A classic example of a buffer is a mixture of acetic acid ( ext{CH}_3 ext{COOH}) and sodium acetate ( ext{CH}_3 ext{COONa}).
  • Acetic acid is weak. It partly dissociates in water, meaning it doesn't give up all its protons easily.
  • Sodium acetate is the conjugate base, which can accept protons released by the acid.
Together, they work well to maintain a consistent pH even when disturbances occur. They are like teammates in a game of pH defense.
pH Calculation
Calculating pH is crucial for understanding the acidity or basicity of a solution. To calculate pH, we often use the Henderson-Hasselbalch equation for buffers: \[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \] Here,
  • \( \text{pH} \) measures how acidic or basic a solution is.
  • \( \text{pKa} \) is the logarithmic measure of the acid dissociation constant. It tells us how strong the acid is.
  • \( [ \text{A}^- ] \) is the concentration of the conjugate base (acetic acid from sodium acetate in this case).
  • \( [ \text{HA} ] \) is the concentration of the weak acid, like acetic acid.
Calculating pH is important in fields like chemistry, biology, and medicine, where maintaining a specific pH is imperative for reactions and processes.
Acetic Acid Dissociation Constant
The dissociation constant \( \text{K}_a \) is a measure of the strength of an acid in solution. For acetic acid, it's denoted by \( 1.8 \times 10^{-5} \). This tells us acetic acid partially dissociates, making it a weak acid. The \( \text{pKa} \) value, derived from \( \text{K}_a \), is a constant for each acid. It's calculated as: \[ \text{pKa} = -\log(\text{K}_a) \] For acetic acid, this calculation gives approximately 4.74.
A smaller \( \text{pKa} \) value indicates a stronger acid, while a higher \( \text{pKa} \) value signifies a weaker acid. Acetic acid鈥檚 \( \text{pKa} \) shows it doesn't dissociate completely, allowing it to function effectively in buffer solutions.
Sodium Acetate in Buffer Solutions
Sodium acetate is a key player in buffer solutions because it supplies acetate ions, the conjugate base of acetic acid. When you dissolve sodium acetate in a solution with acetic acid, you set up a perfect environment to resist changes in pH.Here's how sodium acetate functions in a buffer solution:
  • It dissociates to produce acetate ions \( [\text{CH}_3\text{COO}^-] \).
  • These ions can react with added protons \((\text{H}^+)\), preventing significant pH decreases.
  • Conversely, the acetic acid part of the buffer donates protons \((\text{H}^+)\) when conditions become too basic, maintaining the pH balance.
In practical terms, sodium acetate decides how resistant the buffer is to changes, by determining how available the acetate ions are to soak up excess protons.

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Most popular questions from this chapter

(A): On mixing equal volumes of \(1 \mathrm{M} \mathrm{HCl}\) and \(2 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONa}\), an acidic buffer solution is formed. \((\mathbf{R}):\) Resultant mixture contains \(\mathrm{CH}_{3} \mathrm{COOH}\) and CH, COONa which are parts of acidic buffer.

The oxidation of sulphur dioxide by oxygen to sulphur trioxide has been implicated as an important step in the formation of acid rain: \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) If the equilibrium partial pressures of \(\mathrm{SO}_{2}, \mathrm{O}_{2}\) and \(\mathrm{SO}_{3}\) are \(0.564 \mathrm{~atm}, 0.102 \mathrm{~atm}\), and \(0.333 \mathrm{~atm}\) respec- tively at \(1000 \mathrm{~K}\), what is the \(\mathrm{Kp}\) at that temperature? a. \(2.24\) b. \(4.68\) c. \(3.42\) d. \(13.42\)

\(2.5 \mathrm{~mL}\) of \(2 / 5 \mathrm{M}\) weak monoacidic base \(\left(\mathrm{K}_{\mathrm{b} 1} \times 10^{-12}\right.\) at \(25^{\circ} \mathrm{C}\) ) is titrated with \(2 / 15 \mathrm{M} \mathrm{HCl}\) in water at \(25^{\circ} \mathrm{C}\). The concentration of \(\mathrm{H}+\) at equivalence point is \(\left(\mathrm{Kw}=1 \times 10^{-14}\right.\) at \(\left.25^{\circ} \mathrm{C}\right)\) a. \(3.7 \times 10^{-13} \mathrm{M}\) b. \(3.2 \times 10^{-7} \mathrm{M}\) b. \(3.2 \times 10^{-2} \mathrm{M}\) d. \(2.7 \times 10^{-2} \mathrm{M}\)

Consider the following reversible reaction: \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) \(\Delta \mathrm{H}=-92.5 \mathrm{~kJ}\)Which factor(s) will increase the yield of methanol at equilibrium? a. Addition of inert gas at constant volume b. Increased pressure on the system c. Increased temperature d. Increased partial pressure of hydrogen

Which of the following statement is/are incorrect here? a. Equilibrium constant of an exothermic reaction decreases with the increase of temperature b. \(\mathrm{K}_{\mathrm{p}}\) is always greater than \(\mathrm{K}_{\mathrm{C}}\)
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