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When NaCN dissolves in water, the resulting solution is basic. Account for this observation given that \(\mathrm{p} K_{\mathrm{a}}\) for \(\mathrm{HCN}\) is 9.31.

Short Answer

Expert verified
NaCN solution is basic because CN鈦 hydrolyzes to form OH鈦 and HCN, indicating a basic environment due to high \\text{p}K_b.

Step by step solution

01

Understanding the Dissolution

When sodium cyanide (NaCN) dissolves in water, it dissociates into sodium ions (Na鈦) and cyanide ions (CN鈦). The sodium ions are spectators and do not affect the pH of the solution.
02

Identify the Relevant Ion

The cyanide ion (CN鈦) is crucial for determining the solution's pH. It is capable of interacting with water in a hydrolysis reaction to produce hydroxide ions (OH鈦) and hydrogen cyanide (HCN). The equation for this process is: \ \[ \text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^- \] \ The formation of OH鈦 makes the solution basic.
03

Calculating the Basic Nature Using pKa

The \text{p}K_a of HCN is 9.31, meaning it is a weak acid. Using the relationship \text{p}K_w = \text{p}K_a + \text{p}K_b, where \text{p}K_w is 14, we can calculate \text{p}K_b for CN鈦: \ \[ \text{p}K_b = 14 - 9.31 = 4.69 \] \ This \text{p}K_b value indicates that CN鈦 is a relatively strong base, confirming the solution's basic property.
04

Conclusion

The dissolution of NaCN in water results in a basic solution because CN鈦 acts as a base. It hydrolyzes to form HCN and OH鈦, increasing the solution's pH.

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Key Concepts

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

Sodium Cyanide in Solution
When sodium cyanide (NaCN) is added to water, it dissolves and breaks apart into its constituent ions. These ions are sodium ions (Na鈦) and cyanide ions (CN鈦). While sodium ions typically act as 'spectators' during the reaction, meaning they do not engage in chemical reactions that change the pH, the cyanide ions play a crucial role in influencing the solution's properties. Understanding how these ions behave when in the solution is essential to grasping why the resulting mixture is basic. Essentially, it is the action of CN鈦, rather than Na鈦, that alters the pH and leads to a basic environment.
Understanding Hydrolysis Reactions
The cyanide ion (CN鈦) engages in a hydrolysis reaction when in water. Hydrolysis involves a chemical reaction between an ion and water that produces an acid, a base, or both. In the case of cyanide ions, they react with water to form hydrogen cyanide (HCN) and hydroxide ions (OH鈦).The balanced chemical equation for this hydrolysis reaction is: \[ \text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^- \] The generation of OH鈦 ions increases the basicity of the solution. This transformation is a classic example of how hydrolysis can influence the pH of a mixture by shifting it towards a basic environment.
pKa and pKb in pH Calculations
The relationship between the acid dissociation constant (pKa) and the base dissociation constant (pKb) is key in understanding the chemical behavior of a solution. For any conjugate acid-base pair, the sum of their pKa and pKb is always equal to 14, derived from the product constant of water (pKw).In the case of hydrogen cyanide (HCN) and cyanide ion (CN鈦), the pKa of HCN is 9.31. Thus, the pKb of CN鈦 can be calculated as:\[ \text{p}K_b = 14 - \text{p}K_a = 14 - 9.31 = 4.69 \]This relatively low pKb hints at the cyanide ion's strong basic nature, reinforcing why the solution of NaCN becomes distinctly basic upon dissolution in water.
Properties of Basic Solutions
A basic solution is one that exhibits a pH greater than 7. This occurs when the concentration of hydroxide ions (OH鈦) is greater than that of hydronium ions (H鈦) in the solution. In the case of sodium cyanide in water, the hydrolysis reaction produces OH鈦, which increases the pH and creates a basic environment. Basic solutions are often characterized by a soapy feel and a bitter taste. Understanding the formation of such solutions is important for practical applications in chemistry and industry, where controlling pH is often crucial.
Weak Acids and Bases
Weak acids and bases do not completely dissociate in water. Instead, an equilibrium is established between the undissociated molecules and the ions formed. Like HCN, which is a weak acid with a pKa of 9.31, weak acids only partially release their hydrogen ions. Similarly, weak bases, such as the cyanide ion (CN鈦), partially interact with water to produce hydroxide ions. The equilibrium established contributes to only a moderate increase in pH, characteristic of a basic but not excessively alkaline solution. Comprehending weak acids and bases is essential in recognizing how small changes can significantly affect a solution's pH, crucial for both academic studies and industrial applications.

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Most popular questions from this chapter

Comment on the following observations. (a) In its complexes, Co(III) forms strong bonds to \(O\) - and \(N\) -donor ligands, moderately strong bonds to \(P\) -donor ligands, but only weak bonds to \(A s\) -donor ligands. (b) The values of \(\log K\) for the reaction: \(\mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{X}^{-} \rightleftharpoons[\mathrm{ZnX}]^{+}(\mathrm{aq})\) are 0.7 for \(X=F,-0.2\) for \(X=C 1,-0.6\) for \(X=B r\) and -1.3 for \(X=I\) (c) Phosphine adducts of \(\mathrm{Cr}(\mathrm{III})\) halides can be prepared, but crystallographic studies reveal very long \(\mathrm{Cr}-\mathrm{P}\) bonds (e.g. \(247 \mathrm{pm}\) ).

Four \(\mathrm{p} K_{\mathrm{a}}\) values (1.0,2.0,7.0,9.0) are tabulated for the acid \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}\). Write equations to show the dissociation steps in aqueous solution and assign, with reasoning, a \(\mathrm{p} K_{\mathrm{a}}\) value to each step.

Outline the changes that occur (a) to the salt, and (b) to the water molecules, when solid NaF dissolves in water. How do these changes affect (qualitatively) the entropy of the system?

(a) To what equilibria do the values of \(\mathrm{p} K_{\mathrm{a}}(1)=10.71\) and \(\mathrm{p} K_{\mathrm{a}}(2)=7.56\) for the conjugate acid of \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) refer? (b) Calculate the corresponding values of \(\mathrm{p} K_{\mathrm{b}}\) and write equations to show the equilibria to which these values refer.

Which of the following oxides are likely to be acidic, basic or amphoteric in aqueous solution: (a) \(\mathrm{MgO}\) (b) \(\operatorname{Sn} \mathrm{O} ;(\mathrm{c}) \mathrm{CO}_{2} ;(\mathrm{d}) \mathrm{P}_{2} \mathrm{O}_{5}\) (e) \(\mathrm{Sb}_{2} \mathrm{O}_{3} ;(\mathrm{f}) \mathrm{SO}_{2} ;(\mathrm{g}) \mathrm{Al}_{2} \mathrm{O}_{3}\) (h) \(\mathrm{BeO} ?\)

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