Chapter 6: Problem 67
Calculate the \(\left[\mathrm{OH}^{-}\right]\) in \(0.01 M\) aqueous solution of \(\mathrm{NaOCN}\left(K_{b}\right.\) for \(\left.\mathrm{OCN}^{-}=10^{-10}\right):\) (a) \(10^{-6} \mathrm{M}\) (b) \(10^{-7} M\) (c) \(10^{-8} M\) (d) None of these
Short Answer
Expert verified
The calculated hydroxide ion concentration \(\left[\mathrm{OH}^{-}\right]\) in a 0.01 M aqueous solution of \(\mathrm{NaOCN}\) is \(10^{-6} M\), making option (a) the correct answer.
Step by step solution
01
Understanding the Exercise
We are asked to calculate the hydroxide ion concentration \(\left[\mathrm{OH}^{-}\right]\) in a 0.01 M aqueous solution of sodium cyanate (NaOCN). NaOCN is a salt of a weak acid (HCN) and a strong base (NaOH), so it will hydrolyze in water to form \(\mathrm{OH}^{-}\) ions. The base dissociation constant (Kb) for \(\mathrm{OCN}^{-}\) is given as \(10^{-10}\). The goal is to use this information to calculate \(\left[\mathrm{OH}^{-}\right]\) and see which option (a, b, c) matches the calculated value.
02
Write the Hydrolysis Reaction
The hydrolysis equation for \(\mathrm{OCN}^{-}\) in water can be written as follows: \[\mathrm{OCN}^{-}(aq) + H_2O(l) \rightleftharpoons OH^{-}(aq) + HOCN(aq).\]
03
Determine the Expression for Kb
The expression for the base dissociation constant \(K_b\) of \(\mathrm{OCN}^{-}\) is \[K_b = \frac{\left[\mathrm{OH}^{-}\right]\left[\mathrm{HOCN}\right]}{\left[\mathrm{OCN}^{-}\right]},\] considering the initial concentration of \(\mathrm{OCN}^{-}\) is 0.01 M and will decrease by x, as it reacts with water and subsequently, \(\left[\mathrm{OH}^{-}\right]\) and \(\left[\mathrm{HOCN}\right]\) will increase by x.
04
Use the Approximation for Weak Bases
Since \(K_b\) is very small, it can be assumed that x will also be very small in comparison to 0.01 M. Therefore, the change in concentration of \(\mathrm{OCN}^{-}\) can be ignored and it can be assumed as approx. 0.01 M. This leads to the simplified equation: \[K_b \approx \frac{x^2}{0.01}.\]
05
Calculate the Hydroxide Ion Concentration
Solve for x using the provided value for \(K_b = 10^{-10}\): \[10^{-10} = \frac{x^2}{0.01},\] \[x^2 = 10^{-10} \times 0.01,\] \[x^2 = 10^{-12},\] \[x = \sqrt{10^{-12}},\] \[x = 10^{-6}.\] Thus, the concentration of \(\mathrm{OH}^{-}\) ions, \(\left[\mathrm{OH}^{-}\right]\), is \(10^{-6} M\), which corresponds to option (a).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Base Dissociation Constant (Kb)
The base dissociation constant (\textbf{Kb}) is a measure of the extent to which a base dissociates in water to form hydroxide ions (\textbf{OH\(^{-}\)}) and the corresponding conjugate acid. For a base \textbf{B} in water, the dissociation can be represented as \textbf{B + H\(_2\)O ⇌ OH\(^{-}\) + HB\(^{+}\)}. The value of \(K_b\) is calculated using the expression \[K_b = \frac{[OH^{-}][HB^{+}]}{[B]}\] where [OH\(^{-}\)], [HB\(^{+}\)], and [B] are the equilibrium concentrations of the respective species. A larger \(K_b\) indicates a stronger base, as it suggests a greater degree of dissociation. For the weak base approximation, when the \(K_b\) is small, we assume the change in the concentration of the base (due to dissociation) is negligible, simplifying the calculations for hydroxide ion concentration.
Hydrolysis of Salts
Hydrolysis of salts occurs when salts, formed from the reaction of an acid and base, interact with water to produce either hydronium ions (\textbf{H\(_3\)O\(^{+}\)}) or hydroxide ions (\textbf{OH\(^{-}\)}). This reaction influences the pH of the solution. For example, when a salt derived from a weak acid and a strong base dissolves in water, the anions of the salt will typically undergo hydrolysis, reacting with water to generate hydroxide ions and the corresponding weak acid, as seen with \textbf{NaOCN} in our exercise. \[OCN^{-} + H_2O \leftrightharpoons OH^{-} + HOCN\] Such hydrolysis reactions play a critical role in determining the acidity or basicity of a solution.
Chemical Equilibrium
Chemical equilibrium refers to the state in which the rates of the forward and reverse reactions are equal, resulting in no overall change in the concentrations of reactants and products. This is the point at which a reversible reaction sustains balance. However, equilibrium does not mean the reactants and products are present in equal amounts, but that their concentrations have stabilized at a particular ratio, known as the equilibrium constant (\textbf{K}). In the context of base hydrolysis, our equilibrium constant is \(K_b\), and it represents the equilibrium concentrations of hydroxide ions and the corresponding weak conjugate acid in solution. It is important to understand that while equilibrium describes a steady state, the actual particles are still in motion, continuously reacting in both directions.
Weak Base Approximation
Weak base approximation is a technique used in chemistry to simplify calculations involving weak bases. Under this simplification, we assume that the dissociation of the weak base is so slight that the decrease in its concentration can be considered negligible compared to its initial concentration. This is typically valid when the base dissociation constant (\textbf{Kb}) is very small. In these cases, the change in concentration of the weak base upon dissolving in water is small enough that the initial concentration effectively remains unchanged. This assumption allows us to set up a straightforward equation to solve for the hydroxide ion concentration, as demonstrated in our exercise where the equation \(K_b \approx \frac{x^2}{0.01}\) was used, bypassing the need for more complex equilibrium concentration calculations.