$$ \begin{aligned} &\text { A solution of } 0.01 M \text { concentration of } \mathrm{NH}_{4} \mathrm{OH} \text { is } 2.6 \% \text { dissociated. }\\\ &\text { Calculate }\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right],\left[\mathrm{NH}_{4}^{+}\right],\left[\mathrm{NH}_{4} \mathrm{OH}\right] \text { and } \mathrm{pH} \text { of solution. } \end{aligned} $$

Short Answer

Expert verified
\( \left[\mathrm{OH}^-\right] = \left[\mathrm{NH}_4^+\right] = 2.6 \times 10^{-4}\, M, \left[\mathrm{NH}_4\mathrm{OH}\right] = 9.74 \times 10^{-3}\, M, \left[\mathrm{H}^+\right] = 4.0 \times 10^{-11}\, M, \text{pH} = 10.4 \)

Step by step solution

01

Determine the dissociated concentration of NH4OH

Given that the 0.01 M NH4OH solution is 2.6% dissociated, we can find the concentration of dissociated NH4OH by multiplying the initial concentration (0.01 M) by the percentage of dissociation (2.6%). This will give us the concentration of NH4+ and OH- as NH4OH dissociates into NH4+ and OH- in a 1:1 ratio.\[ \text{Dissociated concentration} = 0.01\, \text{M} \times \frac{2.6}{100} \]
02

Calculate the concentrations of ions

With the dissociated concentration, we can directly write the concentrations of OH- and NH4+ because for every mole of NH4OH that dissociates, one mole of NH4+ and one mole of OH- are produced.\[ \left[\mathrm{OH}^-\right] = \left[\mathrm{NH}_4^+\right] = \text{Dissociated concentration} \]
03

Calculate the remaining concentration of NH4OH

The remaining concentration of NH4OH is the initial concentration minus the dissociated concentration.\[ \left[\mathrm{NH}_4\mathrm{OH}\right] = 0.01\,\text{M} - \text{Dissociated concentration} \]
04

Calculate the concentration of H+ using the water equilibrium constant (Kw)

Knowing OH- concentration, we can find H+ concentration using the equilibrium constant of water (Kw). At 25°C, Kw is \(1.0 \times 10^{-14}\).\[ \left[\mathrm{H}^+\right] = \frac{K_w}{\left[\mathrm{OH}^-\right]} \]
05

Calculate the pH of the solution

The pH is calculated using the concentration of H+ ions.\[ \text{pH} = -\log \left[\mathrm{H}^+\right] \]

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Key Concepts

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

Dissociation in Chemistry
Dissociation in chemistry refers to the process by which molecules or ionic compounds separate into smaller particles, usually ions, in solution. This process is crucial for understanding the behavior of acids, bases, and salts when they dissolve in water. The degree of dissociation indicates how much of the substance has separated into ions.

For example, when a base like ammonia solution (NH4OH) is dissolved in water, it dissociates into ammonium ions (NH4+) and hydroxide ions (OH-). The percentage of dissociation is used to express the ratio of the dissolved substance that has dissociated. In the given exercise, a 2.6% dissociation of 0.01 M NH4OH implies that only a small fraction converts into NH4+ and OH- ions, impacting the overall ionic concentration in the solution.
Ionic Concentration
Ionic concentration, often denoted by square brackets (e.g., \([OH^-]\)), signifies the molar concentration of ions in a solution. Assessing ionic concentrations is essential for predicting the reactivity and properties of the solution, such as conductivity and pH. In the context of weak bases such as NH4OH, not all molecules dissociate to form ions, and the ionic concentration is determined based on the extent of dissociation.

Knowing the original concentration of the weak base and its percentage of dissociation allows us to calculate the specific concentrations of the ions produced. For NH4OH, with 2.6% of the molecules dissociated, the concentrations of NH4+ and OH- can be derived, which are equal due to the 1:1 stoichiometry of the dissociation reaction.
Chemical Equilibrium
Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, leading to a constant ratio of reactants to products in a closed system. It is a dynamic state with ongoing but balanced processes. In the case of weak base dissociation such as NH4OH in water, equilibrium is established between the undissociated base (NH4OH) and the ions produced (NH4+ and OH-).

Equilibrium constants, like the water equilibrium constant (Kw), help us understand the extent of the reaction under certain conditions. Kw, which is \(1.0 \times 10^{-14}\) at 25°C, is the product of the concentrations of H+ and OH- ions in pure water. Calculating ion concentrations in weak base solutions involves using Kw and the known concentrations of other ions to determine unknowns, maintaining the concept of chemical equilibrium.
pOH and pH Relationship
The pH and pOH of a solution are measures of its acidity and basicity, respectively. The pH is the negative logarithm of the hydrogen ion concentration (H+), while pOH is the negative logarithm of the hydroxide ion concentration (OH-). Both scales are logarithmic and inversely related to the concentration of ions in solution.

The relationship between pOH and pH is given by the equation: \(pH + pOH = 14\) at 25°C. This relationship stems from the water equilibrium constant (Kw) and allows us to determine the pH from the pOH (and vice versa). In the textbook exercise, once we calculate the [OH-] concentration, we can find the pOH and then the pH of the weak base NH4OH solution, providing insight into the solution's overall acidity.

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Most popular questions from this chapter

The solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in pure water is \(9.57 \times 10^{-3} \mathrm{~g}\) litre \(^{-1}\). Calculate the pH of its saturated solution. Assume \(100 \%\) ionisation.

$$ \begin{aligned} &\text { The } \mathrm{pH} \text { of } 0.05 \mathrm{M} \text { aqueous solution of diethyl amine is } 12.0 . \text { Calculate }\\\ &K_{\mathrm{b}} \end{aligned} $$

$$ \begin{aligned} &\text { How many mole of } \mathrm{N} \mathrm{H}_{4} \mathrm{Cl} \text { must be added to one litre of } 1.0 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}\\\ &\text { to have a buffer of } \mathrm{pH}=9 . K_{\mathrm{NH}_{4} \mathrm{OH}}=1.8 \times 10^{-5} ? \end{aligned} $$

Benzoic acid is monobasic and when \(1.22 \mathrm{~g}\) of its pure sample after dissolution in distilled water is titrated against a base it uses \(50 \mathrm{~mL}\) of 0.2 \(\mathrm{M} \mathrm{NaOH}\). Calculate the molar mass of benz?ic acid.

Calculate the \(\mathrm{pH}\) of a solution of given mixtures; (a) \(\left(4 \mathrm{~g} \mathrm{CH}_{3} \mathrm{COOH}+6 \mathrm{~g} \mathrm{CH}_{3} \mathrm{COONa}\right)\) in \(100 \mathrm{~mL}\) of mixture; \(K_{\mathrm{a}}\) for \(\mathrm{CH}_{3} \mathrm{COOH}=1.8 \times 10^{-5}\) (b) \(5 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{MOH}+250 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{MCl}\); \(K_{\mathrm{a}}\) for \(M \mathrm{OH}=1.8 \times 10^{-5}\) (c) \(\left(0.25\right.\) mole of \(\mathrm{CH}_{3} \mathrm{COOH}+0.35\) mole of \(\mathrm{CH}_{3} \mathrm{COONa}\) ) in \(500 \mathrm{ml}\) mixture; \(K_{\text {a }}\) for \(\mathrm{CH}_{3} \mathrm{COOH}=3.6 \times 10^{-4}\)

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