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Chapter 16: Problem 104

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
To calculate the pH of the solution made by adding 2.50 g of lithium oxide to enough water to make 1.500 L of solution, you first determine the moles of Li2O dissolved (0.0836 mol), write the balanced equation for the dissociation of Li2O in water (Li2O + H2O -> 2 LiOH), and calculate the moles of OH- ions produced (0.1672 mol). Then, you can find the concentration of OH- ions (0.1115 mol/L) and use the relationship between [H+] and [OH-] to find the concentration of H+ ions (8.97 * 10^{-14} mol/L). Finally, calculate the pH of the solution by taking the negative log of the H+ concentration, resulting in a pH of approximately 13.05.

Step by step solution

01

Calculate the moles of Li2O dissolved

First, we need to convert the 2.50 g of Li2O to moles by using its molar mass, which is given by the sum of the molar masses of 2 lithium atoms and 1 oxygen atom. Molar mass of Li = 6.94 g/mol Molar mass of O = 16.00 g/mol Molar mass of Li2O = (2 * 6.94) + 16.00 = 29.88 g/mol Moles of Li2O = (mass) / (molar mass) = (2.50 g) / (29.88 g/mol) = 0.0836 mol
02

Determine the balanced chemical equation for the dissociation of Li2O in water

Now we need to determine the balanced equation for the dissociation of Li2O in water. Lithium oxide reacts with water to produce lithium hydroxide (LiOH), as shown below: Li2O (s) + H2O (l) -> 2 LiOH (aq) From the balanced equation, we can see that 1 mole of Li2O reacts with 1 mole of water to produce 2 moles of LiOH.
03

Calculate the moles of H+ ions produced

Lithium hydroxide (LiOH) is a strong base, meaning it completely dissociates in water: LiOH (aq) -> Li+ (aq) + OH- (aq) Since 2 moles of LiOH are produced for each mole of Li2O, and each mole of LiOH produces one mole of OH- ions, the number of moles of OH- ions will be double the number of moles of Li2O. Moles of OH- ions = 2 * moles of Li2O = 2 * 0.0836 mol = 0.1672 mol We are now looking for the concentration of H+ ions, so we need to calculate the concentration (in mol/L) of the OH- ions first. This will then allow us to calculate the concentration of the H+ ions using the relationship between the concentrations of H+ and OH- ions: Kw = [H+] [OH-] Where Kw is the ion product of water at 25 掳C, which is equal to 1.0 * 10^{-14}.
04

Calculate the concentration of H+ ions in the solution

First, we need to find the concentration of OH- ions in the solution: [OH-] = moles of OH- ions / volume of solution = 0.1672 mol / 1.500 L = 0.1115 mol/L Now, we can use the relationship between [H+] and [OH-] to find the concentration of H+ ions: Kw = [H+] [OH-] 1.0 * 10^{-14} = [H+] (0.1115 mol/L) Solving for [H+] gives: [H+] = (1.0 * 10^{-14}) / (0.1115 mol/L) = 8.97 * 10^{-14} mol/L
05

Calculate the pH

Finally, we can calculate the pH of the solution using the formula: pH = -log10[H+] pH = -log10(8.97 * 10^{-14}) = 13.05 The pH of the solution is approximately 13.05.

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Most popular questions from this chapter

Predict the products of the following acid-base reactions, and predict whether the equilibrium lies to the left or to the right of the equation: (a) \(\mathrm{O}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) (b) \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{HS}^{-}(a q) \rightleftharpoons\) (c) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\)

Designate the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each of the following equations, and also designate the conjugate acid and conjugate base of each on the right side: (a) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q)\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)\) (c) \(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}{ }^{3-}(a q) \underset{\mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q)}{\rightleftharpoons}\)

\mathrm{~A} 0.100 \mathrm{M}\( solution of bromoacetic acid \)\left(\mathrm{BrCH}_{2} \mathrm{COOH}\right)\( is \)13.2 \%\( ionized. Calculate \)\left[\mathrm{H}^{+}\right],\left[\mathrm{BrCH}_{2} \mathrm{COO}^{-}\right],\left[\mathrm{BrCH}_{2} \mathrm{COOH}\right]\( and \)K_{a}$ for bromoacetic acid.

By what factor does \(\left[\mathrm{H}^{+}\right]\)change for a pH change of (a) \(2.00\) units, (b) \(0.50\) units?

The hypochlorite ion, \(\mathrm{ClO}^{-}\), acts as a weak base. (a) Is \(\mathrm{ClO}^{-}\)a stronger or weaker base than hydroxylamine? (b) When \(\mathrm{ClO}^{-}\) acts as a base, which atom, \(\mathrm{Cl}\) or \(\mathrm{O}\), acts as the proton acceptor? (c) Can you use formal charges to rationalize your answer to part (b)?
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