AP Chemistry Unit 8 Test Review

Buffer solutions containing Na2CO3 and NaHCO3 range in pH from 10.0 to 11.0. The chemical equation below represents the equilibrium between CO32−and H2O, and the table lists the composition of four different buffer solutions at 25°C.

Which of the following chemical equilibrium equations best shows what happens in the buffer solutions to minimize the change in pH when a small amount of a strong base is added?

• H₃O⁺(aq) + OH⁻(aq) ⇌ 2H₂O(l)

• HCO₃⁻(aq) + OH⁻(aq) ⇌ CO₃²⁻(aq) + H₂O(l)

• CO₃²⁻(aq) + H₂O(l) ⇌ HCO₃⁻(aq) + OH⁻(aq)

• CO₃²⁻(aq) + H₃O⁺(aq) ⇌ HCO₃⁻(aq) + H₂O(l)

Answer explanation

The species HCO3− partially consumes the added OH−. Therefore, HCO3−(aq)+OH−(aq)⇄CO32−(aq)+H2O(l) is the chemical equilibrium equation that best shows how the buffer minimizes change in pH when base is added.

Tags

NGSS.HS-PS1-6

Buffer solutions containing Na2CO3 and NaHCO3 range in pH from 10.0 to 11.0. The chemical equation below represents the equilibrium between CO32−and H2O, and the table lists the composition of four different buffer solutions at 25°C.

Which of the following mathematical expressions can be used to determine the approximate pH of buffer 1?

• pH = −log(2.1 x 10⁻⁴) + log(0.100 / 0.150) = 3.50

• pH = −log(2.1 x 10⁻⁴) + log(0.150 / 0.100) = 3.85

• pH = [14.00 + log(2.1 x 10⁻⁴)] + log(0.100 / 0.150) = 10.15

• pH = [14.00 + log(2.1 x 10⁻⁴)] + log(0.150 / 0.100) = 10.50

Answer explanation

The species HCO3− partially consumes the added OH−. Therefore, HCO3−(aq)+OH−(aq)⇄CO32−(aq)+H2O(l) is the chemical equilibrium equation that best shows how the buffer minimizes change in pH when base is added.

Buffer solutions containing Na2CO3 and NaHCO3 range in pH from 10.0 to 11.0. The chemical equation below represents the equilibrium between CO32−and H2O, and the table lists the composition of four different buffer solutions at 25°C.

Which mathematical expression can be used to explain why buffer 2 and buffer 3 have the same pH?

• log(0.200 / 0.200) = log(0.100 / 0.100) = log(1)

• −log(Kₐ) − [ − log(KB)] = pKa

• 0.200 M = 2 x (0.100 M)

• 0.200 M − 0.200 M = 0.100 M − 0.100 M

Answer explanation

Since the solutions are at 25°C, Kb×Ka=Kw, hence pKb+pKa=pKw=14.00. Using pH=pKa+log([A−][HA]) and substituting [14.00+log(2.1×10−4)] for pKw, pH=[14.00+log(2.1×10−4)]+log(0.1000.150)=10.15. This pH is reasonable, since the buffer has a higher concentration of the conjugate acid HCO3− compared with the weak base CO32−; thus, the pH should be slightly lower than 10.32 (compared with buffers 2 and 3).

• The autoionization equilibrium for pure water favors the formation of reactants more at 55 °C than at 25 °C.

• The autoionization equilibrium for pure water produces the same amount of OH⁻ ions at 55 °C and 25 °C.

• At 55 °C, pH = – log (√Kw) for pure water.

• At 55 °C, pH = – log (Kw) for pure water.

Answer explanation

Regardless of the temperature, for pure water Kw=[H3O+][OH−]. At 55°C, [H3O+]=[OH−]=7.0×10−14, and pH =⁢−log(⁢Kw).

Tags

NGSS.HS-PS1-5

NGSS.HS-PS1-6

Based on the information above, which of the following is true for a sample of pure water at 25 °C?

• [H₃O⁺] = 7.0 M

• [OH⁻] = 1.0 x 10⁻¹⁴ M

• pH = 10⁻⁷

• pOH = 7.00

Answer explanation

pOH=−log[OH−], which is equal to −log (1.0×10−7)=7.00 at 25°C.

Tags

NGSS.HS-ESS2-5

The endothermic autoionization of pure water is represented by the chemical equation shown above. The pH of pure water is measured to be 7.00 at 25.0 °C and 6.02 at 100.0 °C. Which of the following statements best explains these observations?

• At the higher temperature water dissociates less, [H₃O⁺] < [OH⁻], and the water becomes basic.

• At the higher temperature water dissociates less, [H₃O⁺] > [OH⁻], and the water remains neutral.

• At the higher temperature water dissociates more, [H₃O⁺] > [OH⁻], and the water becomes acidic.

• At the higher temperature water dissociates more, [H₃O⁺] = [OH⁻], and the water remains neutral.

Answer explanation

Because the reaction is endothermic, increasing temperature will increase Kw. So the water dissociates more, and [H3O+] and [OH−] will always be equal in pure water.

Tags

NGSS.HS-PS1-5

Which of the following gives the best estimate for the pH of a 5 x 10⁻⁴ M Sr(OH)₂ solution at 25 °C?

• pH ≈ 3.0 because Sr(OH)₂ is a strong acid.

• pH ≈ 5.0 because Sr(OH)₂ is a weak acid.

• pH ≈ 9.0 because Sr(OH)₂ is a weak base.

• pH ≈ 11.0 because Sr(OH)₂ is a strong base.

Answer explanation

Sr(OH)2(aq) is a strong base, and its complete dissociation produces 2 moles of OH− ions per mole of Sr(OH)2 that dissolves. As a result, [OH−]=2×(5×10⁻⁴M)=1×10⁻³M and pOH≈3.0, so pH≈11.0.

Which of the following gives the best estimate for the pH of a 1 x 10⁻⁵ M HClO₄(aq) solution at 25 °C?

• pH ≈ 3.0 because HClO₄ is a strong acid.

• pH ≈ 5.0 because HClO₄ is a strong acid.

• pH ≈ 9.0 because HClO₄ is a weak base.

• pH ≈ 11.0 because HClO₄ is a strong base.

Answer explanation

Since HClO4 is a strong acid, it dissociates completely, producing 1 mol of H3O+ per 1 mol of HClO4(aq). Therefore, pH=−log[H3O+]=−log[HClO4]_initial=−log(1×10−5)≈5.0. (In general, the negative log of 10^−x is x. Therefore, the negative log of 1×10−5 is 5.)

Which of the following is the correct mathematical relationship to use to calculate the pH of a 0.10 M aqueous HBr solution?

• pH = [H₃O⁺] = 0.10

• pH = − log(1.0 x 10⁻¹)

• pH = 7.00 − (0.10)

• pH = log (1.0 x 10⁻¹)

Answer explanation

By definition, pH=−log[H3O+]. In this case, pH=−log(1.0×10−1), since for a strong acid like HBr, it is assumed that the acid ionizes 100% and [HBr]_initial=[H3O+].

Tags

NGSS.HS-PS1-7

The equilibrium for the acid ionization of HCOOH is represented by the equation above, and the table gives the percent ionization for HCOOH at different initial concentrations of the weak acid at 25°C. Based on the information, which of the following is true for a 0.125 M aqueous solution of HCOOH?

• It has a higher percent ionization, a lower [H₃O⁺]_eq, and a lower pH than a 0.150 M HCOOH solution does.

• It has a higher percent ionization, a lower [H₃O⁺]_eq, and a higher pOH than a 0.150 M HCOOH solution does.

• It has a lower percent ionization, a higher [H₃O⁺]_eq, and a higher pOH than a 0.100 M HCOOH solution does.

• It has a lower percent ionization, a higher [H₃O⁺]_eq, and a higher pH than a 0.100 M HCOOH solution does.

Answer explanation

Based on the information given, when the initial concentration of the weak acid increases, the percent ionization decreases and the equilibrium concentration of H3O+ increases. A higher [H3O+]eq results in a lower [OH−]eq and a higher pOH.

Tags

NGSS.HS-PS1-5