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The Ka value of formic acid and acetic acid are respectively \(1.77 \times 10^{-4}\) and \(1.75 \times 10^{-5}\). the ratio of the acid strength of \(0.1 \mathrm{~N}\) acids is: (a) \(0.1\) (b) \(0.3\) (c) \(3.178\) (d) 100

Short Answer

Expert verified
The ratio of the acid strength is approximately 10, suggesting a potential mistake in the given options.

Step by step solution

01

Define Acid Strength Ratio

The strength of an acid is proportional to its ionization constant, commonly denoted as \( K_a \). The given problem requires calculating the ratio of the strengths of two acids, formic acid and acetic acid, based on their \( K_a \) values.
02

Identify the Given Values

For formic acid, \( K_{a1} = 1.77 \times 10^{-4} \).For acetic acid, \( K_{a2} = 1.75 \times 10^{-5} \).We need to find the ratio \( \frac{K_{a1}}{K_{a2}} \).
03

Calculate the Ratio

The ratio of the acid strengths is calculated as\[\text{Ratio} = \frac{K_{a1}}{K_{a2}} = \frac{1.77 \times 10^{-4}}{1.75 \times 10^{-5}}\]Divide the two numbers to get:\[\text{Ratio} = \frac{1.77}{1.75} \times \frac{10^{-4}}{10^{-5}} = 1.0114 \times 10^1\]Thus, the ratio is approximately \(10.114 \), which we can round to \(10\).
04

Match with Given Options

From the calculated ratio, approximately \(10\), none of the answer options directly match unless we consider calculation error or improper rounding, as options given are \(0.1, 0.3, 3.178, 100\). Simplification shows a nearest matching choice of \(3.178\) might seem consistent, mistaken between mishandling or misinterpretation in calculations. However, our accurate approximation is \(10\), potentially indicating a discrepancy in provided options.

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Key Concepts

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

Ka value
The \(K_a\) value, or acid dissociation constant, plays a pivotal role in understanding acid strength. It indicates how well an acid can donate its proton (\(H^+\)) in a solution. A higher \(K_a\) value means greater acid strength. This is because the acid dissociates more completely in water, making it a stronger acid. The \(K_a\) value is unique to each acid and is an essential factor when comparing the strengths of different acids.
For instance, in the given exercise, formic acid has a \(K_a\) value of \(1.77 \times 10^{-4}\), whereas acetic acid has a \(K_a\) value of \(1.75 \times 10^{-5}\). From these numbers, you can see that formic acid is stronger, as it has a higher \(K_a\) value. This information then aids in calculating their acid strength ratio by comparing these values.
Ionization constant
The ionization constant, another name for the \(K_a\) value, measures an acid's ability to dissociate in solution. It's a quantitative expression of an acid's strength.
When calculating the ionization constant, the concentration of the products (such as hydrogen ions and the conjugate base) is divided by the concentration of the non-dissociated acid. Depending on the specific values, researchers determine if the acid is weak, moderate, or strong.
The ionization constant remains unchanged at a given temperature, but acids with higher ionization constants tend to be more reactive and stronger due to their higher degree of ionization. That is exactly why in our solution, the ratio gives us a concrete idea of differing strengths between formic and acetic acids.
Formic acid
Formic acid, with the chemical formula \(HCOOH\), is a simple carboxylic acid and is known for being the smallest and simplest of its family. It naturally occurs in ant venom and bee stings.
In chemical terms, formic acid is slightly stronger than many common organic acids. It demonstrates higher acid strength compared to acetic acid, as illustrated by its larger \(K_a\) value of \(1.77 \times 10^{-4}\). This larger \(K_a\) value translates to more effective proton donation to water molecules, making formic acid slightly stronger. It makes it an interesting compound for studying basic principles of organic chemistry and acidity.
Acetic acid
Acetic acid, with the formula \(CH_3COOH\), is widely known as the main component of vinegar that gives it its distinct sour taste. This acid, while common, is weaker than formic acid as shown by its \(K_a\) value of \(1.75 \times 10^{-5}\).
The smaller \(K_a\) value implies less complete ionization in solution, meaning acetic acid donates protons (\(H^+\)) less effectively than formic acid. Understanding this concept is crucial for many applications in cooking, preservation, and production of chemicals. It's an excellent example of how chemical principles directly impact everyday substances.

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Most popular questions from this chapter

In which of the following cases does the reaction go farthest to completion: (a) \(\mathrm{K}=1\) (b) \(\mathrm{K}=10\) (c) \(\mathrm{K}=10^{-2}\) (d) \(\mathrm{K}=10^{2}\)

The relation between \(K_{n}\) and \(K_{c}\) for the reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g})\) is: (a) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{-1}\) (b) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}\) (c) \(K_{p}^{p}=K_{c} /(R T)^{2}\) (d) \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}} / \mathrm{RT}\)

One of the following equilibria is not affected by change in volume of the flask: (a) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g}) \mathrm{Cl}_{2}(\mathrm{~g})\) (b) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) (d) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\)

The equilibrium between water and its vapour, in an open vessel: (a) Can be achieved (b) Depends upon pressure (c) Cannot be achieved (d) Depends upon temperature

For which of the following reaction, \(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}:\) (a) \(2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (b) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) (c) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HCl}(\mathrm{g})\) (d) \(\mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{5}(\mathrm{~g})\)

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