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Chapter 16: Problem 106

Calculate the \(\mathrm{pH}\) of a solution made by adding \(2.50 \mathrm{~g}\) of lithium oxide \(\left(\mathrm{Li}_{2} \mathrm{O}\right)\) to enough water to make \(1.500 \mathrm{~L}\) of solution.

Short Answer

Expert verified
The pH of the solution, made by adding $2.50\:g$ of lithium oxide $\mathrm{Li}_{2} \mathrm{O}$ to enough water to make $1.500\:L$ of solution, is approximately 13.03.

Step by step solution

01

Determine the moles of Li₂O

Before we can start, we need to convert the given mass of lithium oxide to moles. The molar mass of Li₂O can be calculated as follows: Molar mass of Li₂O = \(2\times\) molar mass of Li + molar mass of O = \(2\times 6.94\) g/mol + \(16.00\) g/mol = \(30.88\) g/mol Now, we can calculate the amount in moles: Moles of Li₂O = mass / molar mass = \(2.50\) g / \(30.88\) g/mol ≈ \(0.081\) mol
02

Determine the concentration of LiOH

Lithium oxide (Li₂O) reacts with water to form lithium hydroxide (LiOH) in the following reaction: \[ \mathrm{Li}_{2} \mathrm{O} + \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{LiOH} \] From the stoichiometry of the equation, one mole of Li₂O reacts with one mole of water to produce two moles of LiOH. So, the moles of LiOH produced will be: Moles of LiOH = \(2 \times 0.081\) mol = \(0.162\) mol Since the volume of the solution is 1.500 L, we can now calculate the concentration of LiOH: \[ \mathrm{[LiOH]} = \frac{0.162 \, \text{mol}}{1.5 \, \text{L}} = 0.108 \, \text{M} \]
03

Determine the concentration of OH⁻ ions

Region For every mole of LiOH, one mole of OH⁻ ions is produced, as LiOH is a strong base and dissociates completely in water: \[\mathrm{LiOH} \longrightarrow \mathrm{Li}^+ + \mathrm{OH}^-\] So, the concentration of OH⁻ ions is equal to the concentration of LiOH: \[[OH^-] = 0.108 \, \text{M}\]
04

Calculate the pOH and pH of the solution

Now that we know the concentration of OH⁻ ions, we can calculate the pOH: \[ \mathrm{pOH} = -\log_{10}[\mathrm{OH}^-] = -\log_{10}(0.108) \approx 0.97 \] We then use the following relationship between pH and pOH to find the pH of the solution: \[ \mathrm{pH} + \mathrm{pOH} = 14\] Therefore, \[ \mathrm{pH} = 14 - \mathrm{pOH} = 14 - 0.97 \approx 13.03 \]
05

Final Answer

The pH of the solution is approximately 13.03.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(0.0167 \mathrm{M} \mathrm{HNO}_{3},\) (b) \(0.225 \mathrm{~g}\) of \(\mathrm{HClO}_{3}\) in \(2.00 \mathrm{~L}\) of solution, (c) \(15.00 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\) diluted to \(0.500 \mathrm{~L}, (\mathrm{~d})\) a mixture formed by adding \(50.0 \mathrm{~mL}\) of \(0.020 \mathrm{M} \mathrm{HCl}\) to \(125 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{HI}\)

Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) dimethylamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH} ;\) (b) carbonate ion, \(\mathrm{CO}_{3}^{2-} ;(\mathbf{c})\) formate ion, \(\mathrm{CHO}_{2}^{-}\).

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for (a) \(1.5 \times 10^{-3} \mathrm{M} \mathrm{Sr}\left(\mathrm{OH}_{2}\right),\) (b) \(2.250 \mathrm{~g}\) of \(\mathrm{LiOH}\) in \(250.0 \mathrm{~mL}\) of solution, (c) \(1.00 \mathrm{~mL}\) of \(0.175 \mathrm{M} \mathrm{NaOH}\) diluted to \(2.00 \mathrm{~L},\) (d) a solution formed by adding \(5.00 \mathrm{~mL}\) of \(0.105 \quad M \mathrm{KOH}\) to \(15.0 \mathrm{~mL}\) of \(9.5 \times 10^{-2} \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\)

Indicate whether each of the following statements is correct or incorrect. For those that are incorrect, explain why they are wrong. (a) Every Bronsted-Lowry acid is also a Lewis acid. (b) Every Lewis acid is also a Bronsted-Lowry acid. (c) Conjugate acids of weak bases produce more acidic solutions than conjugate acids of strong bases. (d) \(\mathrm{K}^{+}\) ion is acidic in water because it causes hydrating water molecules to become more acidic. (e) The percent ionization of a weak acid in water increases as the concentration of acid decreases.

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 \(\mathrm{ClO}^{-},\) (b) \(\mathrm{BrO}^{-}\) or \(\mathrm{BrO}_{2}^{-},(\mathbf{c})\) \(\mathrm{HPO}_{4}^{2-}\) or \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)
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