According to the Bronsted-Lowry definition, an acid is a substance that does which of the following?
Donates a proton (H+)
Accepts a proton (H+)
Donates an electron pair
Increases the OH- concentration
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WhyA Bronsted-Lowry acid is a proton (H+) donor, while a Bronsted-Lowry base is a proton acceptor.
2Multiple choice · Easy
A solution has [H+] = 1.0 x 10^-4 M at 25 degrees C. What is the pH of the solution?
4.0
10.0
-4.0
1.0
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WhypH = -log[H+] = -log(1.0 x 10^-4) = 4.0.
3Multiple choice · Easy
At 25 degrees C, a solution with pH = 9 is best described as which of the following?
Basic
Acidic
Neutral
Saturated
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WhyA pH greater than 7 indicates a basic solution; pH less than 7 is acidic and pH = 7 is neutral at 25 degrees C.
4Multiple choice · Easy
Which of the following is classified as a strong acid that ionizes essentially completely in water?
HCl
CH3COOH
HF
HCN
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WhyHCl is a strong acid that dissociates nearly 100% in water; CH3COOH, HF, and HCN are weak acids that only partially ionize.
5Fill in the blank · Easy
At 25 degrees C, the ion-product constant of water, Kw, equals [H+][OH-] = 1.0 x 10^-14. For any aqueous solution at this temperature, the sum pH + pOH equals .
Answer:
14 / 14.0
WhySince Kw = 1.0 x 10^-14 at 25 degrees C, taking -log of both sides gives pH + pOH = 14.
6Fill in the blank · Easy
For a weak acid HA, the equilibrium expression for its ionization constant is Ka = ([H+][A-]) / .
Answer:
[HA] / HA
WhyFor HA <-> H+ + A-, Ka equals the product of product concentrations divided by the concentration of the undissociated acid [HA].
7Multiple choice · Medium
What is the pH of a 0.025 M solution of the strong base NaOH at 25 degrees C?
12.40
1.60
11.60
12.00
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WhyNaOH fully dissociates, so [OH-] = 0.025 M; pOH = -log(0.025) = 1.60, and pH = 14.00 - 1.60 = 12.40.
8Multiple choice · Medium
A 0.10 M solution of a weak acid HA has a pH of 3.0 at 25 degrees C. What is the approximate Ka of HA?
1.0 x 10^-5
1.0 x 10^-3
1.0 x 10^-6
1.0 x 10^-7
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Why[H+] = 10^-3 = 1.0 x 10^-3 M, so Ka = (1.0 x 10^-3)^2 / (0.10 - 0.001) approximately = 1.0 x 10^-6 / 0.10 = 1.0 x 10^-5.
9Multiple choice · Medium
Which statement correctly describes the relationship between an acid's Ka and the Kb of its conjugate base at 25 degrees C?
Ka x Kb = Kw = 1.0 x 10^-14
Ka + Kb = Kw = 1.0 x 10^-14
Ka / Kb = Kw = 1.0 x 10^-14
Ka = Kb for all conjugate pairs
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WhyFor a conjugate acid-base pair, Ka x Kb = Kw = 1.0 x 10^-14, so a stronger acid (larger Ka) has a weaker conjugate base (smaller Kb).
10Multiple choice · Medium
A buffer is prepared with equal concentrations of CH3COOH (Ka = 1.8 x 10^-5) and CH3COO-. What is the approximate pH of this buffer at 25 degrees C?
4.74
9.26
7.00
5.26
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WhyWhen [acid] = [base], the Henderson-Hasselbalch equation gives pH = pKa = -log(1.8 x 10^-5) = 4.74.
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