What are the major species present in a 0.150-M \(\mathrm{NH}_{3}\) solution? Calculate the \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of this solution.

Short Answer

Expert verified
The major species present in a 0.150 M NH\(_{3}\) solution are NH\(_{3}\) and H\(_{2}\)O. The concentration of OH\(^{-}\) ions in the solution is calculated to be \(1.64 \times 10^{-3}\) M. The pH of this solution is found to be 11.216.

Step by step solution

01

Identify the major species in the solution

The major species present in a 0.150 M NH\(_{3}\) solution are: 1. NH\(_{3}\) (ammonia, the solute) 2. H\(_{2}\)O (water, the solvent)
02

Write the equilibrium expression for the dissociation of NH\(_{3}\) in water

When NH\(_{3}\) reacts with water, it acts as a base, accepting a proton (H\(^{+}\)) from water to form the ammonium ion, NH\(_{4}^{+}\), and the hydroxide ion, OH\(^{-}\). The chemical reaction and the resulting equilibrium expression are: NH\(_{3}\) + H\(_{2}\)O \(\rightleftharpoons\) NH\(_{4}^{+}\) + OH\(^{-}\) \(K_b = \frac{[\text{NH}_{4}^{+}][\text{OH}^{-}]}{[\text{NH}_{3}]}\) Here, \(K_b\) is the base dissociation constant for NH\(_{3}\).
03

Find the value of \(K_b\) for NH\(_{3}\)

Using a reference table or textbook, find the \(K_b\) value for NH\(_{3}\). The base dissociation constant for NH\(_{3}\) is: \(K_b = 1.8 \times 10^{-5}\)
04

Calculate the equilibrium concentrations of NH\(_{4}^{+}\) and OH\(^{-}\)

Let \(x\) represent the concentration of NH\(_{4}^{+}\) and OH\(^{-}\) produced at equilibrium. Therefore, the change in NH\(_{3}\) concentration is also \(x\). The equilibrium concentrations can be represented as follows: [NH\(_{4}^{+}\)] = [OH\(^{-}\)] = x [NH\(_{3}\)] = 0.150 - x Now, substitute these concentrations into the \(K_b\) expression and solve for \(x\): \(K_b = 1.8 \times 10^{-5} = \frac{x \cdot x}{0.150 - x}\) Since \(K_b\) is very small, the amount of dissociation is minimal, and we can approximate \(0.150 - x \approx 0.150\). Therefore: \(1.8 \times 10^{-5} = \frac{x^2}{0.150}\)
05

Calculate the [OH\(^{-}\)] concentration

Solve for \(x\), which represents the equilibrium [OH\(^{-}\)] concentration: \(x = \sqrt{1.8 \times 10^{-5} \times 0.150} = \sqrt{2.7\times 10^{-6}}\) [OH\(^{-}\)] = \(1.64 \times 10^{-3}\) M
06

Calculate the pH of the solution

Since we have found the concentration of OH\(^{-}\), we can now calculate the pOH and subsequently find the pH of the solution: pOH = -log([OH\(^{-}\)]) = -log(1.64 × 10\(^{-3}\)) pOH = 2.784 Next, use the relationship between pH and pOH: pH + pOH = 14 Now, solve for pH: pH = 14 - pOH = 14 - 2.784 = 11.216 So, the pH of the 0.150 M NH\(_3\) solution is 11.216.

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Most popular questions from this chapter

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