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Chapter 14: Problem 132

Will the following oxides give acidic, basic, or neutral solutions when dissolved in water? Write reactions to justify your answers. a. \(\mathrm{Li}_{2} \mathrm{O}\) b. \(\mathrm{CO}_{2}\) c. \(\mathrm{SrO}\)

Short Answer

Expert verified
a. Li₂O + H₂O ⟶ 2 LiOH; forms a basic solution (LiOH is a strong alkali). b. CO₂ + H₂O ⟶ H₂CO₃; forms an acidic solution (H₂CO₃ is a weak acid). c. SrO + H₂O ⟶ Sr(OH)₂; forms a basic solution (Sr(OH)₂ is a strong alkali).

Step by step solution

01

Reaction of Li₂O with water

First, we will write the balanced reaction for lithium oxide (Li₂O) dissolving in water (H₂O). Li₂O + H₂O ⟶ 2 LiOH The resulting product is lithium hydroxide (LiOH), which is a strong alkali. Since the solution contains a hydroxide ion (OH⁻), it will be a basic solution.
02

Reaction of CO₂ with water

Next, we will write the balanced reaction for carbon dioxide (CO₂) dissolving in water (H₂O). CO₂ + H₂O ⟶ H₂CO₃ The resulting product is carbonic acid (H₂CO₃), which is a weak acid. Since the solution contains an acidic component (H+), it will be an acidic solution.
03

Reaction of SrO with water

Finally, we will write the balanced reaction for strontium oxide (SrO) dissolving in water (H₂O). SrO + H₂O ⟶ Sr(OH)₂ The resulting product is strontium hydroxide (Sr(OH)₂), which is a strong alkali. Since the solution contains a hydroxide ion (OH⁻), it will be a basic solution. In conclusion: a. Lithium oxide (Li₂O) forms a basic solution when dissolved in water. b. Carbon dioxide (CO₂) forms an acidic solution when dissolved in water. c. Strontium oxide (SrO) forms a basic solution when dissolved in water.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of a \(0.050 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(1.4 \times 10^{-5}\).

A \(10.0\) -mL sample of an \(\mathrm{HCl}\) solution has a \(\mathrm{pH}\) of \(2.000\). What volume of water must be added to change the \(\mathrm{pH}\) to \(4.000\) ?

A certain acid, HA, has a vapor density of \(5.11 \mathrm{~g} / \mathrm{L}\) when in the gas phase at a temperature of \(25^{\circ} \mathrm{C}\) and a pressure of \(1.00 \mathrm{~atm}\). When \(1.50 \mathrm{~g}\) of this acid is dissolved in enough water to make \(100.0 \mathrm{~mL}\) of solution, the \(\mathrm{pH}\) is found to be \(1.80\). Calculate \(K_{\mathrm{a}}\) for the acid.

A solution is prepared by adding \(50.0 \mathrm{~mL}\) concentrated hydrochloric acid and \(20.0 \mathrm{~mL}\) concentrated nitric acid to \(300 \mathrm{~mL}\) water. More water is added until the final volume is \(1.00 \mathrm{~L}\). Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\), and the \(\mathrm{pH}\) for this solution. [Hint: Concentrated HCl is \(38 \%\) HCl (by mass) and has a density of \(1.19\) \(\mathrm{g} / \mathrm{mL} ;\) concentrated \(\mathrm{HNO}_{3}\) is \(70 . \% \mathrm{HNO}_{3}\) (by mass) and has a density of \(1.42 \mathrm{~g} / \mathrm{mL} .\) ]

Consider \(1000 . \mathrm{mL}\) of a \(1.00 \times 10^{-4} M\) solution of a certain acid HA that has a \(K_{\mathrm{a}}\) value equal to \(1.00 \times 10^{-4} .\) How much water was added or removed (by evaporation) so that a solution remains in which \(25.0 \%\) of HA is dissociated at equilibrium? Assume that HA is nonvolatile.
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