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\(0.005 \mathrm{M}\) acid solution has \(5 \mathrm{pH}\). The percentage ionization of acid is: (a) \(0.8 \%\) (b) \(0.6 \%\) (c) \(0.4 \%\) (d) \(0.2 \%\)

Short Answer

Expert verified
The percentage ionization of the acid is 0.2%, option (d).

Step by step solution

01

Understanding pH and Hydrogen Ion Concentration

pH is the measure of the hydrogen ion concentration in a solution. It can be calculated using the formula: \[ \text{pH} = - \log[\text{H}^+] \]Given that the solution has a pH of 5, we can find the hydrogen ion concentration by rearranging the formula:\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-5} \text{ M} \]
02

Calculating Concentration of Ionized Acid

Since the \[ [\text{H}^+] = 10^{-5} \text{ M} \], it means that this is the concentration of the ionized molecules of the acid. The initial concentration of the acid is given as \(0.005 \text{ M}\).
03

Determining the Percentage Ionization

The percentage ionization is given by:\[ \text{Percentage Ionization} = \left( \frac{\text{Concentration of Ionized Acid}}{\text{Initial Concentration of Acid}} \right) \times 100 \]Substitute the known values:\[ \text{Percentage Ionization} = \left( \frac{10^{-5} \text{ M}}{0.005 \text{ M}} \right) \times 100 \]\[ \text{Percentage Ionization} = \left( \frac{1 \times 10^{-5}}{5 \times 10^{-3}} \right) \times 100 \]\[ \text{Percentage Ionization} = \left( \frac{1}{500} \right) \times 100 \approx 0.2\% \]
04

Matching the Percentage Ionization to Options

The calculated percentage ionization is \(0.2\%\), which matches option (d).

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Key Concepts

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

Percentage Ionization
Percentage ionization is a handy concept when studying acids. It tells us what fraction of the acid molecules ionize or dissociate in a solution. This value is expressed as a percentage, making it easier to compare the ionization level of different acids in similar conditions. To compute the percentage ionization, we follow a simple formula:
  • Identify the concentration of ionized acid, which often equals the concentration of hydrogen ions (H^+) in the solution.
  • Determine the initial concentration of the acid provided before any dissociation.
  • Apply the formula: Percentage Ionization = \( \left( \frac{\text{Ionized Acid}}{\text{Initial Acid Concentration}} \right) \times 100 \)
This calculation helps in understanding how strong or weak an acid is. A strong acid will have a higher percentage ionization, nearly 100%, while a weak acid will have a lower percentage, often significantly below 100%.
Hydrogen Ion Concentration
Understanding hydrogen ion concentration is a key part of pH chemistry. The concentration of hydrogen ions (H^+) in a solution directly influences the pH of that solution. The term pH is simply a scale that helps us gauge the acidity or basicity of a solution using the formula:\[ \text{pH} = - \log[\text{H}^+] \]When you have a pH value, you can calculate the H^+ concentration by rearranging the equation:\[ [\text{H}^+] = 10^{-\text{pH}} \]For instance, a solution with a pH of 5 will have a hydrogen ion concentration of \( 10^{-5} \text{ M} \). This means the lower the pH, the higher the H+ concentration, indicating a more acidic solution. This relationship is vital for chemistry studies as it also impacts how substances behave in reactions.
Acid Concentration
Acid concentration refers to the amount of acid present in a solution, usually expressed in molarity (M). It's important for determining how the acid will behave in reactions, how it impacts pH, and how much it ionizes in a solution. There are a few key points to remember about acid concentration:
  • The higher the concentration, the more acid molecules are present in the solution, which can contribute to a lower pH.
  • In calculations, such as those for ionization, the initial concentration is critical to find how much of the acid dissociates into ions.
  • A concentrated acid means it's potent and potentially more reactive, whereas a dilute acid has fewer acid molecules and is often less reactive.
Understanding acid concentration is a cornerstone in pH chemistry, aiding students in predicting how acids will behave and react in various situations.

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Most popular questions from this chapter

If the equilibrium constant for the reaction, \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at 750 is 49 , then the equilibrium constant for the reaction, \(\mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons 1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+3 / 2 \mathrm{H}_{2}(\mathrm{~g})\) at the same temperature will be: (a) \(1 / 49\) (b) 49 (c) \(1 / 7\) (d) \(49^{2}\)

In what manner will increase of pressure affect the following equation: \(\mathrm{C}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g})\) (a) Shift in the reverse direction (b) Shift in the forward direction (c) Increase in the yield of hydrogen (d) No effect

\(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) in the above reaction \(K_{p}\) and \(K_{c}\) are related as: (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} \times(\mathrm{RT})^{-1}\) (c) \(K_{c}=K_{p} \times(R T)^{2}\) (d) \(K_{p}=K_{c} \times(R T)^{-2}\)

One mole of \(\mathrm{A}(\mathrm{g})\) is heated to \(300^{\circ} \mathrm{C}\) in a closed one litre vessel till the following equilibrium is reached. \(\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})\) The equilibrium constant of this reaction at \(300^{\circ} \mathrm{C}\) is 4\. What is the concentration of \(\mathrm{B}\) (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) at equilibrium: (a) \(0.2\) (b) \(0.6\) (c) \(0.75\) (d) \(0.1\)

For a gaseous equilibrium: \(2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}), \mathrm{K}_{\mathrm{p}}\) has a value of \(1.8\) at \(700 \mathrm{~K} .\) What is the value of \(\mathrm{K}_{\mathrm{c}}\) for the equilibrium \(2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~A}\) at the same pressure: (a) \(0.031\) (b) \(1.3 \times 10^{-3}\) (c) \(44.4\) (d) 38

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