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Chapter 1: Problem 58

(a) One acid has a \(\mathrm{p} K_{\mathrm{a}}\) of 2 , the other has a \(\mathrm{p} K_{\mathrm{a}}\) of 8 . What is the ratio of their \(K_{\mathrm{a}}\) 's? (b) Two acids differ by a factor of 10,000 in their \(K_{\mathrm{a}}^{\prime} \mathrm{s}\). If the \(\mathrm{p} K_{\mathrm{a}}\) of the weaker acid is 5 , what is the \(\mathrm{p} K_{\mathrm{a}}\) of the stronger acid?

Short Answer

Expert verified
(a) The ratio of Ka's is 1,000,000. (b) The pKa of the stronger acid is 1.

Step by step solution

01

Understand the relationship between pKa and Ka

Recall that pKa and Ka are related through the formula: \[pK_a = -\log_{10}(K_a)\]This means that if you know the pKa values, you can calculate the Ka values using inverse logarithm properties.
02

Calculate Ka values from given pKa values

For the first acid (pKa = 2), calculate Ka:\[pK_a_1 = 2 \Rightarrow K_a_1 = 10^{-2} = 0.01\]For the second acid (pKa = 8), calculate Ka:\[pK_a_2 = 8 \Rightarrow K_a_2 = 10^{-8} = 0.00000001\]
03

Determine the ratio of their Ka values

To find the ratio of the Ka values:\[\text{Ratio} = \frac{K_a_1}{K_a_2} = \frac{0.01}{0.00000001} = 10^6 = 1,000,000\]Thus, the ratio of the Ka values is 1,000,000.
04

Use the factor difference to find pKa of the stronger acid

Given that the acids differ by a factor of 10,000:\[K_{a_{weaker}} = 10^{-5}, \quad K_{a_{stronger}} = 10,000 \times K_{a_{weaker}}\]\[K_{a_{stronger}} = 10,000 \times 10^{-5} = 10^{-1}\]Now find the pKa of the stronger acid:\[pK_{a_{stronger}} = -\log_{10}(10^{-1}) = 1\]
05

Summarize the results

For part (a), the ratio of their Ka's is 1,000,000. For part (b), the pKa of the stronger acid is 1.

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Key Concepts

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

pKa calculation
The concept of pKa is central to understanding acid-base equilibria. pKa represents the negative logarithm of the acid dissociation constant, Ka. In simpler terms, pKa is a way to express acidity. The lower the pKa, the stronger the acid, meaning the acid dissociates more completely in water. Understanding this relationship is crucial when comparing the strength of different acids. The formula connecting pKa and Ka is: \[ pK_a = -\log_{10}(K_a) \] This logarithmic relationship tells us that a small difference in pKa translates into a large difference in acidity. For instance, an acid with a pKa of 2 is considerably stronger than one with a pKa of 8. This is because, as you will see in the sections that follow, each single unit change in pKa corresponds to a tenfold change in the acid's Ka value.
Ka ratio
The ratio of two acids' Ka values is a straightforward calculation once you understand the pKa values. In the original problem, you were given two acids with pKa values of 2 and 8. To find the ratio of their Ka's: - Convert the pKa values into Ka using the formula: \[ K_a = 10^{-pK_a} \] - So, for pKa = 2: \[ K_a = 10^{-2} = 0.01 \] - And for pKa = 8: \[ K_a = 10^{-8} = 0.00000001 \] The ratio of their Ka values is the division of these two numbers: \[ \text{Ratio} = \frac{K_{a_{1}}}{K_{a_{2}}} \] Thus, you find: \[ \frac{0.01}{0.00000001} = 10^6 = 1,000,000 \] Clearly, the first acid is 1,000,000 times stronger in terms of dissociation than the second acid.
logarithmic relationship in chemistry
Logarithmic relationships are common in chemistry, particularly when dealing with pH, pKa, and other logarithmically scaled measurements. The logarithm functions as a tool to simplify the numbers and relationships we deal with. For acids, pKa is a more manageable way of expressing the acidic strength. To interpret these logarithmic relationships:
  • pKa = -log(Ka) transforms the Ka value into a simpler scale.
  • Each whole number difference in pKa indicates a tenfold difference in hydrogen ion concentration.
  • For example, moving from a pKa of 5 to a pKa of 6 means the acid is 10 times weaker.
In the given problem, the drastic shift in Ka values over a pKa range illustrates just how powerful logarithms can be in expressing the massive variations in acid strength that can result from seemingly small changes. Such mathematical relationships make it easier for chemists to visualize and understand these differences.

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Most popular questions from this chapter

Write an equation for the Lewis acid/Lewis base reaction between boron trifluoride and each of the following. Use curved arrows to track the flow of electrons and show formal charges if present. (a) Fluoride ion (b) Dimethyl sulfide \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S}\right]\) (c) Trimethylamine \(\left[\left(\mathrm{CH}_{3}\right)_{3} N\right]\) Sample Solution (a) Fluoride ion has 8 electrons (4 pairs) in its valence shell. It acts as a Lewis base and uses one pair to bond to boron in \(\mathrm{BF}_{3}\).

Write Lewis formulas, including unshared pairs, for each of the following. Carbon has four bonds in each compound. (a) Propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) (c) Methyl fluoride \(\left(\mathrm{CH}_{3} \mathrm{~F}\right)\) (b) Methanol \(\left(\mathrm{CH}_{4} \mathrm{O}\right)\) (d) Ethyl fluoride \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{~F}\right)\) Sample Solution (a) The Lewis formula of propane is analogous to that of ethane but the chain has three carbons instead of two. The ten covalent bonds in the Lewis formula shown account for 20 valence electrons, which. is the same as that calculated from the molecular formula \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\). The eight hydrogens of \(\mathrm{C}_{3} \mathrm{H}_{8}\) contribute 1 electron each and the three carbons 4 each, for a total of \(20(8\) from the hydrogens and 12 from the carbons). Therefore, all the valence electrons are in covalent bonds; propane has no unshared pairs.

Indicate the direction of the dipole for the following bonds using the symbol \(\longrightarrow\) and \(\delta+\), \(\delta\) - notation. \(\begin{array}{llllll}\mathrm{H}-0 & \mathrm{H}-\mathrm{N} & \mathrm{C}-0 & \mathrm{C}=0 & \mathrm{C}-\mathrm{N} & \mathrm{C}=\mathrm{N}\end{array}\)

Referring to the periodic table as needed, write electron configurations for all the elements in the third period. Sample Solution The third period begins with sodium and ends with argon. The atomic number \(Z\) of sodium is 11 , and so a sodium atom has 11 electrons. The maximum number of electrons in the 1s, \(2 \mathrm{~s}\), and \(2 \mathrm{p}\) orbitals is ten, and so the eleventh electron of sodium occupies a 3 s orbital. The electron configuration of sodium is \(1 s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{2} 2 p_{z}^{2} 3 s^{1}\).

Which is the stronger base in each of the following pairs? (Note: This information will prove useful when you get to Chapter 9.) (a) Sodium ethoxide \(\left(\mathrm{NaOCH}_{2} \mathrm{CH}_{3}\right)\) or sodium amide \(\left(\mathrm{NaNH}_{2}\right)\) (b) Sodium acetylide \((\mathrm{NaC} \equiv \mathrm{CH})\) or sodium amide \(\left(\mathrm{NaNH}_{2}\right)\) (c) Sodium acetylide \((\mathrm{NaC} \equiv \mathrm{CH})\) or sodium ethoxide \(\left(\mathrm{NaOCH}_{2} \mathrm{CH}_{3}\right)\) Sample Solution (a) \(\mathrm{NaOCH}_{2} \mathrm{CH}_{3}\) contains the ions \(\mathrm{Na}^{+}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-} . \mathrm{NaNH}_{2}\) contains the ions \(\mathrm{Na}^{+}\) and \(\mathrm{H}_{2} \mathrm{~N}^{-} . \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\) is the conjugate base of ethanol; \(\mathrm{H}_{2} \mathrm{~N}^{-}\) is the conjugate base of ammonia. \(\begin{array}{lll}\text { Base } & \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-} & \mathrm{H}_{2} \mathrm{~N}^{-} \\ \text {Conjugate acid } & \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} & \mathrm{NH}_{3} \\ \mathrm{p} K_{\mathrm{a}} \text { of conjugate acid } & 16 & 36\end{array}\) The conjugate acid of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\) is stronger than the conjugate acid of \(\mathrm{H}_{2} \mathrm{~N}^{-}\). Therefore, \(\mathrm{H}_{2} \mathrm{~N}^{-}\) is a stronger base than \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\).
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