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Chapter 14: Problem 95

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

Short Answer

Expert verified
The major species present in a 0.150 M NH3 solution are NH3, NH4+, and OH-. The concentration of OH- ions is 1.64 x 10^-3 M, and the pH of the solution is approximately 11.22.

Step by step solution

01

Write the chemical equilibrium for the reaction

Write the chemical reaction for ammonia's interaction with water. NH3 (aq) + H2O (l) <=> NH4+ (aq) + OH- (aq)
02

Write the Kb expression for ammonia

The Kb expression for ammonia (NH3) is given by: Kb = \(\frac{[NH_4^+][OH^-]}{[NH_3]}\) The Kb value for ammonia (NH3) is 1.8 x 10^-5. We will use this value later when we solve for the concentration of OH- ions.
03

Set up an ICE table

An ICE (Initial, Change, Equilibrium) table will help us keep track of the concentrations of each species throughout the reaction. Since we are given that the initial concentration of NH3 is 0.150 M, our table will look like this: NH3 + H2O <=> NH4+ + OH- Initial: 0.150 - 0 0 Change: -x - x x Equilibrium: 0.150-x - x x
04

Substitute ICE table values into the Kb expression

Substitute the equilibrium expressions from the ICE table into the Kb expression we wrote in Step 2: \(1.8 \times 10^{-5} = \frac{x^2}{0.150 - x}\) Since Kb is relatively small, we can make a simplifying assumption that x is also small and the concentration of NH3 will not change significantly: \(1.8 \times 10^{-5} = \frac{x^2}{0.150}\)
05

Solve for x, which represents [OH-]

Now, we will solve for x, which represents the concentration of OH-: \(x^2 = 1.8 \times 10^{-5} \times 0.150\) \(x^2 = 2.7 \times 10^{-6}\) \(x = \sqrt{2.7 \times 10^{-6}}\) \(x = 1.64 \times 10^{-3}\) So, the concentration of OH-, [OH-], is 1.64 x 10^-3 M.
06

Calculate the pH of the solution

Now that we have the concentration of hydroxide ions, [OH-], we can use it to compute for the pOH, and then find the pH of the solution. We will use the following formula: \(pOH = -\log{[OH^{-}]}\) Then we'll use the relationship between pH and pOH: pH + pOH = 14 First, calculate the pOH: \(pOH = -\log{(1.64 \times 10^{-3})}\) \(pOH = 2.78\) Next, calculate the pH: pH = 14 - pOH = 14 - 2.78 = 11.22 The pH of the 0.150 M NH3 solution is approximately 11.22.

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Most popular questions from this chapter

Making use of the assumptions we ordinarily make in calculating the \(\mathrm{pH}\) of an aqueous solution of a weak acid, calculate the pH of a \(1.0 \times 10^{-6}-\mathrm{M}\) solution of hypobromous acid \(\left(\mathrm{HBrO}, K_{\mathrm{a}}=2 \times 10^{-9}\right) .\) What is wrong with your answer? Why is it wrong? Without trying to solve the problem, explain what has to be included to solve the problem correctly.

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