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Chapter 1: Problem 62

Consider \(1.0 \mathrm{M}\) aqueous solutions of each of the following. Which solution is more basic? (a) Sodium cyanide \((\mathrm{NaCN})\) or sodium fluoride (NaF) (b) Sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right)\) or sodium acetate (c) Sodium sulfate \(\left(\mathrm{Na}_{2} \mathrm{SO}_{4}\right)\) or sodium methanethiolate \(\left(\mathrm{NaSCH}_{3}\right)\)

Short Answer

Expert verified
(a) Sodium cyanide; (b) Sodium carbonate; (c) Sodium methanethiolate

Step by step solution

01

Understanding Basicity

Basicity in a solution is often related to the ability of the anion of the salt to accept protons or produce hydroxide ions in water. An anion derived from a weak acid is more basic, while anions from strong acids are not.
02

Examine Sodium Cyanide (NaCN) and Sodium Fluoride (NaF)

NaCN dissociates into Na鈦� and CN鈦�. CN鈦� is derived from HCN, a weak acid, making it a strong base. NaF dissociates into Na鈦� and F鈦�, with F鈦� from HF, a weak acid, but weaker than HCN. So, CN鈦� is more basic than F鈦�.
03

Compare Sodium Carbonate (Na鈧侰O鈧�) and Sodium Acetate

Na鈧侰O鈧� dissociates into Na鈦� and CO鈧兟测伝, where CO鈧兟测伝 can form HCO鈧冣伝, indicating strong basicity as it is derived from H鈧侰O鈧�, a weak acid. Sodium acetate dissociates into Na鈦� and CH鈧僀OO鈦�, where CH鈧僀OO鈦� is derived from acetic acid, a weak acid but stronger than carbonic acid. Therefore, CO鈧兟测伝 is more basic than CH鈧僀OO鈦�.
04

Evaluate Sodium Sulfate (Na鈧係O鈧�) and Sodium Methanethiolate (NaSCH鈧�)

Na鈧係O鈧� dissociates into Na鈦� and SO鈧劼测伝. Sulfate is derived from H鈧係O鈧�, a strong acid, making it a weak base. Sodium methanethiolate dissociates into Na鈦� and SCH鈧冣伝, which is derived from methanethiol (a weak acid). Therefore, SCH鈧冣伝 is more basic than SO鈧劼测伝.

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Key Concepts

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

Anion Basicity
In aqueous solutions, basicity is determined by the ability of an anion to accept protons or produce hydroxide ions \((\text{OH}^{-})\). This ability is linked to the strength of the parent acid from which the anion is derived.
  • Anions from weak acids are generally more basic because they more readily accept protons, thus raising the solution's pH.
  • Anions derived from strong acids do not exhibit significant basicity, as they are stable in their deprotonated form.
When examining anions in solutions, identifying whether their parent acid is weak or strong gives insight into their basicity. Examples include:
  • CN鈦�: From the weak acid HCN.
  • F鈦�: From the weak acid HF.
  • CO鈧兟测伝: From H鈧侰O鈧�, another weak acid.
  • CH鈧僀OO鈦�: From acetic acid, which is weak but stronger than H鈧侰O鈧�.
Sodium Cyanide vs Sodium Fluoride
Sodium cyanide (\(\text{NaCN}\)) and sodium fluoride (\(\text{NaF}\)) both dissociate in water, forming different anions with distinct basicity levels.
  • The cyanide ion (CN鈦�)is derived from hydrogen cyanide (HCN), a particularly weak acid, making CN鈦籥 relatively strong base.
  • Fluoride ion (F鈦�)is derived from hydrofluoric acid (HF), which, while also weak, is a stronger acid than HCN.
Thus, CN鈦� is more basic than F鈦籨ue to its origin from a weaker acid. As a result, solutions of \(\text{NaCN}\) tend to be more basic than those of \(\text{NaF}\).Understanding these differences helps predict the resulting pH of their respective solutions and allows comparisons of their basic nature.
Sodium Carbonate vs Sodium Acetate
Sodium carbonate (\(\text{Na}_2\text{CO}_3\)) and sodium acetate break down into CO鈧兟测伝and CH鈧僀OO鈦籭ons respectively, each influencing the basicity of their solutions.
  • Carbonate ion (CO鈧兟测伝)is derived from carbonic acid (H鈧侰O鈧�), a weak acid, allowing CO鈧兟测伝to act as a strong base by converting to bicarbonate (HCO鈧冣伝)in water.
  • The acetate ion (CH鈧僀OO鈦�)comes from acetic acid, a weak acid but one stronger than carbonic acid.
Thus, CO鈧兟测伝is more basic than CH鈧僀OO鈦籨ue to the lower acidity of its parent acid, making \(\text{Na}_2\text{CO}_3\)supplies more alkaline than \(\text{NaCH}_3\text{COO}\).Comprehending this difference is key to recognizing why sodium carbonate solutions often yield a higher pH.
Sodium Sulfate vs Sodium Methanethiolate
When analyzing sodium sulfate (\(\text{Na}_2\text{SO}_4\)) and sodium methanethiolate (\(\text{NaSCH}_3\)), we need to consider their anions: sulfate and methanethiolate.
  • Sulfate ion (SO鈧劼测伝)is associated with sulfuric acid (H鈧係O鈧�), a strong acid, which leaves SO鈧劼测伝with minimal basicity since it's stable and doesn't accept protons easily.
  • In contrast, methanethiolate ion (SCH鈧冣伝)is derived from methanethiol, a weak acid, making SCH鈧冣伝a far stronger base compared to SO鈧劼测伝.
Therefore, solutions of \(\text{NaSCH}_3\)exhibit more pronounced basic characteristics than those of \(\text{Na}_2\text{SO}_4\).This is essential for predicting the behavior of these solutions in reactive environments.

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Most popular questions from this chapter

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

Which compound in each of the following pairs would you expect to have the greater dipole moment \(\mu\) ? Why? (a) \(\mathrm{HF}\) or \(\mathrm{HCl}\) (d) \(\mathrm{CHCl}_{3}\) or \(\mathrm{CCl}_{3} \mathrm{~F}\) (b) \(\mathrm{HF}\) or \(\mathrm{BF}_{3}\) (e) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) or \(\mathrm{CH}_{3} \mathrm{OH}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CH}\) or \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) (f) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) or \(\mathrm{CH}_{3} \mathrm{NO}_{2}\)

Write structural formulas for all the constitutionally isomeric compounds having the given molecular formula. (a) \(\mathrm{C}_{4} \mathrm{H}_{10}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) (e) \(\mathrm{C}_{3} \mathrm{H}_{9} \mathrm{~N}\) (b) \(\mathrm{C}_{5} \mathrm{H}_{12}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br}\)

Which is a stronger base, ethoxide \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \ddot{\mathrm{O}}:^{-}\right)\) or tert-butoxide \(\left[\mathrm{CH}_{3}\right)_{3} \mathrm{CO}:\) ]?

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 \(1 s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{2} 2 p_{z}^{2} 3 s^{1}\).
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