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Chapter 16: Problem 100

A solution is made by adding \(0.300 \mathrm{g} \mathrm{Ca}(\mathrm{OH})_{2}(s), 50.0 \mathrm{mL}\) of \(1.40 \mathrm{MNN}_{3},\) and enough water to make a final volume of 75.0 \(\mathrm{mL}\) . Assuming that all of the solid dissolves, what is the pH of the final solution?

Short Answer

Expert verified
The pH of the final solution when 0.300 g of Ca(OH)₂, 50.0 mL of 1.40 M NH₃, and enough water to make the final volume 75.0 mL is approximately 13.034.

Step by step solution

01

Calculate the initial moles of Ca(OH)₂ and NH₃ in the solution

To do this, we will use the given information: - mass of Ca(OH)₂ = 0.300 g - volume of NH₃ solution = 50.0 mL - concentration of NH₃ = 1.40 M First, we need to convert the mass of Ca(OH)₂ to moles. We will use the molar mass of Ca(OH)₂ which is about 74.1 g/mol. Moles of Ca(OH)₂ = mass / molar mass Moles of Ca(OH)₂ = \(0.300 \mathrm{g} / 74.1 \mathrm{g/mol}\) Moles of Ca(OH)₂ ≈ 0.00405 mol Next, we will calculate the moles of NH₃ by multiplying its concentration by the volume of the solution: Moles of NH₃ = concentration × volume (in L) Moles of NH₃ = 1.40 M × 0.050 L Moles of NH₃ = 0.0700 mol
02

Calculate concentrations in final solution

Now that we have the initial moles of both Ca(OH)₂ and NH₃, we will find their concentrations in the final solution. The total volume of the final solution is 75.0 mL: Concentration of Ca(OH)₂ = moles / volume Concentration of Ca(OH)₂ = \(0.00405 \mathrm{mol} / 0.075 \mathrm{L}\) Concentration of Ca(OH)₂ ≈ 0.0540 M Concentration of NH₃ = moles / volume Concentration of NH₃ = \(0.0700 \mathrm{mol} / 0.075 \mathrm{L}\) Concentration of NH₃ ≈ 0.933 M
03

Calculate concentration of OH⁻ ions

Ca(OH)₂ dissociates in water to form 2 moles of OH⁻ ions for every 1 mole of Ca(OH)₂. Therefore, the concentration of OH⁻ ions in the solution will be: Concentration of OH⁻ = 2 × concentration of Ca(OH)₂ Concentration of OH⁻ = 2 × 0.0540 M Concentration of OH⁻ ≈ 0.108 M
04

Calculate the pOH of the solution

Now calculate the pOH of the solution using the concentration of OH⁻ ions: pOH = - log₁₀([OH⁻]) pOH = - log₁₀(0.108) pOH ≈ 0.966
05

Calculate the pH of the solution

To find the pH, we will use the relationship between pH and pOH: pH + pOH = 14 pH = 14 - pOH pH = 14 - 0.966 pH ≈ 13.034 The pH of the final solution is approximately 13.034.

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Most popular questions from this chapter

At the freezing point of water \(\left(0^{\circ} \mathrm{C}\right), K_{w}=1.2 \times 10^{-15}\) Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) for a neutral solution at this temperature.

Predict the stronger acid in each pair: (a) HNO \(_{3}\) or HNO \(_{2};\) \((\mathbf{b})\mathrm{H}_{2} \mathrm{S}\) or \(\mathrm{H}_{2} \mathrm{O} ;(\mathbf{c}) \mathrm{H}_{2} \mathrm{SO}_{4}\) or \(\mathrm{H}_{2} \mathrm{SeO}_{4}(\mathbf{d}) \mathrm{CH}_{3} \mathrm{COOH}\) or \(\mathrm{CCl}_{3}\) \(\mathrm{COOH}.\)

Based on their compositions and structures and on conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) \(\mathrm{BrO}^{-}\) or \(\mathrm{ClO}^{-},(\mathbf{b}) \mathrm{BrO}^{-}\) or \(\mathrm{BrO}_{2}^{-}\) (c) \(\mathrm{HPO}_{4}^{2-}\) or \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}.\)

Addition of phenolphthalein to an unknown colorless solution does not cause a color change. The addition of bromthymol blue to the same solution leads to a yellow color. (a) Is the solution acidic, neutral, or basic? (b) Which of the following can you establish about the solution: (i) A minimum pH, (ii) A maximum pH, or (iii) A specific range of pH values? (c) What other indicator or indicators would you want to use to determine the pH of the solution more precisely?

Calculate \(\left[\mathrm{OH}^{-}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: \((\mathbf{a})\left[\mathrm{H}^{+}\right]=0.0505 M (\mathbf{b})\left[\mathrm{H}^{+}\right]=2.5 \times 10^{-10} M ;(\mathbf{c})\) a solution in which \(\left[\mathrm{H}^{+}\right]\) is 1000 times greater than \(\left[\mathrm{OH}^{-}\right] .\)
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