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Chapter 6: Problem 86

What is the \(\mathrm{pH}\) of a solution of \(0.28 \mathrm{M}\) acid and \(0.84 \mathrm{M}\) of its conjugate base if the ionization constant of acid is \(4 \times 10^{-4}\) ? (a) \(3.88\) (b) \(3.34\) (c) 7 (d) \(10.12\)

Short Answer

Expert verified
The pH of the solution is 3.88.

Step by step solution

01

Identify the Relevant Equation

For the given acid-base mixture, the Henderson-Hasselbalch equation is applied to find the pH of the buffer solution. The equation is given by: \[ \text{pH} = \text{p}K_a + \log \left( \frac{\text{[A^-]}}{\text{[HA]}} \right) \]
02

Calculate the pKa

Calculate the pKa using the given ionization constant (Ka) of the acid. The pKa is the negative logarithm of Ka: \[ \text{p}K_a = -\log(K_a) \]\[ \text{p}K_a = -\log(4 \times 10^{-4}) \]
03

Compute the pKa Value

Using a calculator to find the log of the Ka value:\[ \text{p}K_a = -\log(4 \times 10^{-4}) \approx 3.40 \]
04

Use the Henderson-Hasselbalch Equation

Insert the pKa value and concentrations of the acid and its conjugate base into the Henderson-Hasselbalch equation to calculate pH: \[ \text{pH} = 3.40 + \log \left( \frac{0.84}{0.28} \right) \]
05

Calculate the log ratio

Now compute the log of the ratio of the concentrations of the conjugate base ([A^-]) and the acid ([HA]): \[ \log \left( \frac{0.84}{0.28} \right) = \log(3) \approx 0.48 \]
06

Determine the pH of the Solution

Add the pKa value to the log ratio to find the pH:\[ \text{pH} = 3.40 + 0.48 = 3.88 \]

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Key Concepts

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

pH Calculation
Understanding the pH of a solution is crucial as it determines how acidic or basic that solution is. pH is measured on a scale from 0 to 14 where a pH less than 7 indicates acidity, a pH of 7 is neutral, and a pH above 7 points towards basicity. The pH can be calculated using the Henderson-Hasselbalch equation when dealing with buffer solutions. This relates the pH, pKa (which is a measure of acid strength), and the ratio of the concentrations of the conjugate base to the acid in solution. One key insight is that a greater concentration of conjugate base will increase the pH, signaling a more basic solution.
Acid-Base Buffer Solution
A buffer solution is a special solution that resists changes in pH upon the addition of small amounts of either acid or base. It usually consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. Buffers work due to the equilibrium established between these pairs; when excess H+ ions are added, the base component neutralizes them, and when OH- ions are added, the acid component reacts with them. The ability of a buffer to maintain pH is dependent on the presence of both components in sufficient quantities. This buffering capacity is quantified in the Henderson-Hasselbalch equation, which helps to calculate the pH of buffer solutions efficiently.
Ionization Constant
The ionization constant, represented as Ka, is a value that tells us how well an acid dissociates in water. It's based on the equilibrium concentration of the acid and its dissociated ions. A larger Ka indicates a stronger acid that dissociates more completely. The ionization constant is also involved when calculating the pH of solutions that contain acids by using the Henderson-Hasselbalch equation. It is important to note that the ionization constant is temperature-dependent and varies with different acids, making it crucial for accurate pH calculations.
pKa Calculation
pKa is another important term in acid-base chemistry. It is the negative base-10 logarithm of the ionization constant (Ka). This value helps us understand the strength of an acid; the smaller the pKa, the stronger the acid, because it implies a greater tendency for the acid to donate its proton to water. When calculating the pKa from the Ka value, we take the negative log of Ka, which provides a more manageable number to work with, especially when applying it in the Henderson-Hasselbalch equation to determine the pH of solutions. Calculating pKa is a crucial step in predicting the behavior of acids and bases under a variety of chemical conditions.

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Most popular questions from this chapter

Number of equivalents of \(\mathrm{HCl}\) present in \(100 \mathrm{~mL}\) of its solution whose \(\mathrm{pH}\) is \(4:\) (a) \(10^{-4}\) (b) \(10^{-3}\) (c) \(10^{-2}\) (d) \(10^{-5}\)

\(1.0 L\) solution is prepared by mixing \(61 \mathrm{gm}\) benzoic acid \(\left(p K_{a}=4.2\right)\) with \(72 \mathrm{gm}\) of sodium benzoate and then \(300 \mathrm{~mL} 1.0 \mathrm{M}\) HBr solution was added. The \(\mathrm{pH}\) of final solution is: (a) \(3.6\) (b) \(3.8\) (c) \(4.2\) (d) \(4.8\)

A hand book states that the solubility of \(R \mathrm{NH}_{2}(g)\) in water at 1 atm and \(25^{\circ} \mathrm{C}\) is \(22.41\) volumes of \(R \mathrm{NH}_{2}(g)\) per volume of water. \(\left(\mathrm{pK}_{b}\right.\) of \(R \mathrm{NH}_{2}=4\) ) Find the max. pOH that can be attained by dissolving \(\mathrm{RNH}_{2}\) in water : (a) 1 (b) 2 (c) 4 (d) 6 .

Calculate the \(\left[\mathrm{OH}^{-}\right]\) in \(0.01 M\) aqueous solution of \(\mathrm{NaOCN}\left(K_{b}\right.\) for \(\left.\mathrm{OCN}^{-}=10^{-10}\right):\) (a) \(10^{-6} \mathrm{M}\) (b) \(10^{-7} M\) (c) \(10^{-8} M\) (d) None of these

To a \(10 \mathrm{~mL}\) of \(10^{-3} \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution water has been added to make the total volume of one litre. Its pOH would be : (a) 3 (b) 12 (c) 9 (d) 5
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