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The percentage hydrolysis of \(\mathrm{NaCN}\) in \(\left(\frac{\mathrm{N}}{80}\right)\) aqueous solution [Dissociation constant of \(\mathrm{HCN}\) is \(1.3 \times 10^{-9}\) and \(\left.\mathrm{K}_{\mathrm{w}}=1.0 \times 10^{-14}\right]\) is (a) \(8.2\) (b) \(9.6\) (c) \(5.26\) (d) \(2.48\)

Short Answer

Expert verified
The percentage hydrolysis of NaCN is 2.48%, option (d).

Step by step solution

01

Understand the problem

We're asked to calculate the percentage hydrolysis of sodium cyanide (NaCN) in the given concentration of solution. We are provided with the dissociation constant of hydrogen cyanide (HCN) and the ionic product of water (Kw).
02

Determine the initial concentration of NaCN

Given the solution's concentration is \( \frac{N}{80} \), meaning normality is \( \frac{1}{80} \) N. For NaCN (a salt from a strong base and a weak acid), the initial concentration of CN^- ions is also \( \frac{1}{80} \) M (since it completely dissociates).
03

Use the hydrolysis formula for anion derived from a weak acid

For the hydrolysis of anion \( \text{CN}^- \):\[ K_h = \frac{K_w}{K_a} \]where \( K_a \) is the dissociation constant of HCN and \( K_w \) is given. Calculate \( K_h \):\[ K_h = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-9}} = 7.692 \times 10^{-6} \]
04

Calculate the degree of hydrolysis

Using the degree of hydrolysis formula based on concentration:\[ h = \sqrt{K_h / C} \]where \( C \) is the concentration of the solution.\[ h = \sqrt{7.692 \times 10^{-6} \times 80} \]
05

Simplify the hydrolysis calculation

Substitute the values into the formula:\[ h = \sqrt{7.692 \times 10^{-6} \times 80} = \sqrt{0.00061536} = 0.0248 \]
06

Calculate percentage hydrolysis

Convert the degree of hydrolysis to a percentage:\[ \text{Percentage hydrolysis} = h \times 100 = 0.0248 \times 100 = 2.48 \]
07

Choose the correct option

Based on the calculation, the percentage hydrolysis is 2.48%, hence the correct answer is (d) 2.48.

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Key Concepts

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

Sodium Cyanide and its Hydrolysis
Sodium cyanide (NaCN) is a chemical compound composed of sodium and cyanide ions. In water, NaCN dissociates completely to release sodium (Na鈦) and cyanide (CN鈦) ions. This characteristic is typical of salts formed from strong bases, like sodium hydroxide, and weak acids, such as hydrogen cyanide (HCN). When sodium cyanide undergoes hydrolysis in water, the CN鈦 ion reacts with H鈧侽 to form HCN and OH鈦 ions.
  • This reaction shows why the solution is basic: CN鈦 accepts protons from water, forming OH鈦, which increases the pH of the solution.
  • Understanding the extent of this hydrolysis and how it affects the pH is essential for various industrial and chemical processes.
The ability of CN鈦 to form HCN in water is central to calculating its hydrolysis, which is an integral part of solving related chemistry problems effectively.
Understanding Dissociation Constant (Ka)
The dissociation constant, denoted as Ka, is an important parameter when dealing with weak acids. It measures the strength of an acid in solution, indicating the degree to which it can donate protons (H鈦). For hydrogen cyanide (HCN), a weak acid, the dissociation constant is given as 1.3 x 10鈦烩伖. This small value reflects HCN's weak tendency to donate protons in water.
  • When calculating hydrolysis, Ka is used to determine the extent of anionic reactions in solution.
  • Lower Ka values typically indicate stronger tendency towards hydrolysis for the corresponding anion, like CN鈦 in water.
In the context of sodium cyanide鈥檚 hydrolysis, a precise understanding of Ka helps in accurately applying formulas that predict the chemical behavior of dilute solutions.
The Ionic Product of Water (Kw)
The ionic product of water, represented as Kw, is a fundamental constant in chemistry that defines the product of the concentrations of hydrogen ions (H鈦) and hydroxide ions (OH鈦) in pure water. At 25掳C, Kw is typically 1.0 x 10鈦宦光伌. This value is crucial in understanding the equilibrium state of water and aqueous solutions.
  • Kw plays a key role in calculating the degree of hydrolysis, since it reflects the ion balance that occurs when acids or bases are added to water.
  • Understanding Kw helps predict the behavior of other ions in solution, especially in the case of weak acids and bases.
Accurate use of Kw allows for the calculation of important equilibrium constants like Kh, ensuring the correct application of the hydrolysis formula.
Degree of Hydrolysis and Its Calculation
Degree of hydrolysis (h) represents how much of a compound like sodium cyanide undergoes hydrolysis in solution. It is the fraction of CN鈦 that reacts with water to form HCN and OH鈦 ions. Calculating the degree of hydrolysis involves the concentration of the solution and the hydrolysis constant.Using the formula:

Degree of Hydrolysis Formula

\[ h = \sqrt{\frac{K_w}{K_a} \times C} \]
  • This formula finds the extent of hydrolysis by considering the initial concentration (C) of the solution and balances the opposing actions of ion breakdown.
  • The percentage hydrolysis is derived by multiplying h by 100, offering a clear measure of the reaction scale.
A high degree of hydrolysis indicates greater conversion of cyanide to hydrogen cyanide, thus impacting the solution's pH and its practical applications.

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Most popular questions from this chapter

If \(50 \mathrm{ml}\) of \(0.2 \mathrm{M} \mathrm{NaCN}\) is mixed with \(50 \mathrm{ml}\) of \(0.2\) M \(\mathrm{HCl}\), then \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{CN}^{-}\right]=\mathrm{x} \times 10^{-6}\) where \(\mathrm{x}\) is \(\left(\mathrm{K}_{\mathrm{b}}\right.\) for \(\left.\mathrm{CN}^{-}=2 \times 10^{-5}\right)\)

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If molar concentrations of two weak acids are the same, their relative strengths can be compared by (a) \(\frac{\alpha_{1}}{\alpha_{2}}\) (b) \(\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}\) (c) \(\sqrt{\mathrm{K}_{1} / \mathrm{K}_{2}}\) (d) \(\frac{\left[\mathrm{H}^{+}\right]_{1}}{\left[\mathrm{H}^{+}\right]_{2}}\)

How many litres of water must be added to litre of an aqueous solution of \(\mathrm{HCl}\) with a \(\mathrm{pH}\) of 1 to create an aqueous solution with \(\mathrm{pH}\) of 2 ? [2013] (a) \(2.0 \mathrm{~L}\) (b) \(9.0 \mathrm{~L}\) (c) \(0.1 \mathrm{~L}\) (d) \(0.9 \mathrm{~L}\)

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