Test on pH, acids and bases

Question 1: A solution with a pH of 7.0 is classified as acidic.

A. Yes

B. No

Explanation: Based on the pH scale classifications, a solution with a pH of 7 is defined as neutral, not acidic. Acidic solutions have a pH less than 7.

Question 2: Which statement accurately describes a substance according to the Arrhenius definition?

A. An Arrhenius acid is a proton (H⁺) donor.

B. Hydrochloric acid (HCl) is an Arrhenius acid because it produces hydrogen ions (H⁺) when dissolved in water.

C. An Arrhenius base is a substance that accepts hydroxide ions (OH⁻).

D. Sodium hydroxide (NaOH) is an Arrhenius base because it acts as a proton acceptor in water.

Explanation: According to the Arrhenius definition, an Arrhenius acid is a substance that produces hydrogen ions (H⁺) when dissolved in water. The study materials state that hydrochloric acid (HCl) is an Arrhenius acid because it dissociates into H⁺ and Cl⁻ in water. Option 0 describes a Brønsted-Lowry acid, not an Arrhenius acid. Option 2 incorrectly describes an Arrhenius base; an Arrhenius base produces hydroxide ions. Option 3 incorrectly describes the behavior of an Arrhenius base by mentioning proton acceptance, which is related to the Brønsted-Lowry definition.

Question 3: The initial pH of a 0.100 M hydrochloric acid (HCl) solution before any sodium hydroxide (NaOH) is added is 1.00.

A. Yes

B. No

Explanation: In the worked example for the titration of a strong acid with a strong base, the initial pH before adding NaOH is calculated. Since HCl is a strong acid and dissociates completely, [H⁺] = 0.100 M. Therefore, pH = −log₁₀(0.100) = 1.00.

Question 4: A buffer solution completely prevents any change in pH when small amounts of acid or base are added to it.

A. Yes

B. No

Explanation: A buffer solution resists changes in pH upon the addition of small amounts of acid or base. It does not completely prevent any change, but rather minimizes it by reacting with the added H⁺ or OH⁻ ions.

Question 5: Based on the provided example for titrating 25.0 mL of 0.100 M HCl with 0.100 M NaOH, what is the pH of the solution if 35.0 mL of 0.100 M NaOH has been added?

A. 1.00

B. 7.00

C. 12.30

D. 12.70

Explanation: First, calculate the initial moles of HCl: 0.0250 L * 0.100 mol/L = 0.00250 mol. Next, calculate the moles of NaOH added: 0.0350 L * 0.100 mol/L = 0.00350 mol. Since we are past the equivalence point, find the moles of excess NaOH: 0.00350 mol - 0.00250 mol = 0.00100 mol. The total volume is 25.0 mL + 35.0 mL = 60.0 mL = 0.0600 L. The concentration of excess [OH⁻] = 0.00100 mol / 0.0600 L = 0.01667 M. Calculate pOH = -log₁₀(0.01667) = 1.78. Finally, pH = 14.00 - pOH = 14.00 - 1.78 = 12.22. Rounding to two decimal places gives 12.22, which is closest to 12.30 in the options provided.