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Calculate the \(\mathrm{pH}\) of a \(0.2 \mathrm{M} \mathrm{NH}_{3}\) solution for which \(\mathrm{K}_{\mathrm{b}}=1.8 \times 10^{-5}\) at \(25^{\circ} \mathrm{C}\). The equation for the reaction is \(\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightleftarrows \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-}\)

Short Answer

Expert verified
The pH of the 0.2 M NH鈧 solution with a Kb = 1.8 x 10鈦烩伒 is approximately 8.083.

Step by step solution

01

Write the given information

We are provided with the following information: Concentration of ammonia (NH鈧) = 0.2 M Kb for ammonia = 1.8 x 10鈦烩伒 We also know the equation for the reaction: NH鈧 + H鈧侽 鈬 NH鈧勨伜 + OH鈦
02

Set up the Kb expression

To calculate the concentration of hydroxide ions, we need to write the Kb expression. For the given reaction, the Kb expression is: Kb = \(\frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]}\)
03

Construct the table of ion concentrations

Now let's create a table of ion concentrations before, during and at equilibrium. Let x be the concentration of NH鈧勨伜 and OH鈦 ions at equilibrium. | Species | Initial Concentration (M) | Change (M) | Equilibrium Concentration (M) | |--- |--- |--- |--- | | NH鈧 | 0.2 | -x | 0.2-x | | NH鈧勨伜 | 0 | +x | x | | OH鈦 | 0 | +x | x |
04

Substituting concentrations into the Kb expression

Substitute the equilibrium concentrations into the Kb expression: \(1.8\times10^{-5}=\frac{x^2}{0.2-x}\)
05

Solve for x, the concentration of OH鈦 ions

In this step, we will solve for x, the concentration of OH鈦 ions at equilibrium. Since Kb is very small, we can assume x to be negligible in comparison to 0.2 M. Therefore, the equation becomes: \(1.8\times10^{-5}=\frac{x^2}{0.2}\) Rearrange and solve for x: \(x=\sqrt{1.8\times10^{-5}\times0.2}\) \(x=1.2094\times10^{-6}\) M This is the equilibrium concentration of OH鈦 ions.
06

Calculate pOH

For the next step, we will calculate the pOH. The pOH can be obtained using the formula: pOH = -log鈧佲個 [OH鈦籡 Plug in the value of x (the equilibrium concentration of OH鈦 ions) into the formula: pOH = -log鈧佲個 (1.2094 x 10鈦烩伓) = 5.917
07

Calculate pH

Finally, we need to calculate the pH of the solution. We can find this using the relationship between pH and pOH: pH + pOH = 14 So, to find the pH: pH = 14 - pOH = 14 - 5.917 = 8.083 The pH of the 0.2 M NH鈧 solution is approximately 8.083.

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Key Concepts

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

Equilibrium Concentration
Understanding equilibrium concentration is crucial for calculating the pH of solutions in chemical reactions. Equilibrium is achieved in a reversible reaction when the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of the reactants and products remain constant.

In our context, while calculating the pH of the ammonia solution, we acknowledge that the initial concentration of ammonia solutions may change until equilibrium is established. Using the dissociation equation, we establish the concentrations of each species during equilibrium. These are called equilibrium concentrations. Once at equilibrium, we can apply equilibrium expressions, like Kb in this scenario, to solve for unknown concentrations such as that of hydroxide ions \([OH^-]\).
Ammonia Solution
An ammonia solution, often referred to as aqueous ammonia, is a mixture of ammonia (NH鈧) and water. Ammonia is a weak base, which means it does not fully ionize in water to produce hydroxide ions \(OH^-\). Instead, it establishes an equilibrium between the ammonia itself, ammonium ions \(NH_4^+\), and hydroxide ions.

The equilibrium reaction of ammonia with water is represented by:
  • NH鈧 + H鈧侽 鈬 NH鈧勨伜 + OH鈦
This equation shows that ammonia absorbs a proton from water, converting into the ammonium ion and releasing hydroxide ions. The quantity of hydroxide ions produced impacts the pH of the solution, making the solution slightly basic.

This is why a calculation of pH needs consideration of the base nature of ammonia, as well as its dissociation constant.
Equilibrium Table
An equilibrium table is a helpful tool to track the concentration changes of reactants and products as a chemical reaction goes to equilibrium. Using an initial-change-equilibrium (ICE) table can simplify the solution calculation in equilibrium problems.

In our step-by-step solution for ammonia, the equilibrium table helps visualize the given initial concentrations and track changes as the system reaches equilibrium. This understanding allows us to write the equilibrium expression accurately, like Kb, using the known quantities.

As seen in the example:
  • Reactant: NH鈧, initial concentration of 0.2 M decreases as NH鈧 dissociates.
  • Products: NH鈧勨伜 and OH鈦 gain concentration \((+x)\) as they are formed.
The equilibrium table provides essential detail, allowing precise determination of concentrations at equilibrium, which is crucial for further calculations like pH.
Base Dissociation Constant
The base dissociation constant, or Kb, measures the extent of ionization of a base in solution. It's a specific equilibrium constant for bases, similar to how Ka is for acids.

Given by the formula:
  • \[ K_{b} = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} \]
This expression quantitatively describes the ability of ammonia to produce OH鈦 the ions in water. A smaller Kb value, like the one for ammonia \((1.8 \times 10^{-5})\), indicates a weaker base, meaning NH鈧 only partially ionizes in water.

Understanding and applying Kb in calculations helps determine the solution's equilibrium concentrations. It's crucial in predicting whether a solution will act more as a base or remain more neutral in a chemical setting.

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Most popular questions from this chapter

A chemist mixes \(.5\) moles of acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) and \(.5\) moles of HCN with enough water to make a one liter solution. Calculate the final concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}\) and \(\mathrm{OH}^{-}\). Assume the following constants: \(\mathrm{K}_{\mathrm{HC} 2 \mathrm{H} 3 \mathrm{O} 2}=1.8 \times 10^{-5}, \mathrm{~K}_{\mathrm{HCN}}=4 \times 10^{-10}, \mathrm{~K}_{\mathrm{W}}=1.0 \times 10^{-14}\)

Find the \(\mathrm{pH}\) of a \(0.1 \mathrm{M}\) solution of ammonia, \(\mathrm{NH}_{3} \cdot \mathrm{pK}_{\mathrm{b}}=\) \(1.76 \times 10^{-5}\)

A chemist mixes together \(100 \mathrm{ml}\) of \(0.5 \mathrm{M}\) sodium acetate, \(100 \mathrm{ml}\) of \(0.25 \mathrm{M}\) hydrochloric acid \((\mathrm{HCl})\), and \(100 \mathrm{ml}\) of a \(1.0 \mathrm{M}\) salt solution. She dilutes this to \(1000 \mathrm{ml}\). Determine the concentrations of all the ions present, undissociated acetic acid, and the final \(\mathrm{pH}\) of the solution, using \(\mathrm{K}_{\mathrm{a}}\) of acetic acid \(=1.7 \times 10^{-5}\) and \(\mathrm{K}_{\mathrm{W}}=1.0 \times 10^{-14}\).

At normal body temperature, \(37^{\circ} \mathrm{C}\left(98.6^{\circ} \mathrm{F}\right)\), the ionization constant of water, \(\mathrm{K}_{\mathrm{w}}\), is \(2.42 \times 10^{-14}\) moles \(^{2} /\) liter \(^{2}\). A physician injects a neutral saline solution into a patient. What will be the \(\mathrm{pH}\) of this solution when it has come into thermal equilibrium with the patient's body?

Find the \(\mathrm{pH}\) of \(0.10 \mathrm{M}\) HOAc solution that has \(0.20 \mathrm{M} \mathrm{NaOAc}\) dissolved in it. The dissociation constant of HOAc is \(1.75 \times 10^{-5}\)

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