Acid/Base M.C AP Chem

Consider the Ka values for the following acids: Cyanic acid, HOCN, 3.5 x10-4 Formic acid, HCHO2, 1.7 x 10-4 Lactic acid, HC3H5O3, 1.3 x 10-4 Propionic acid, HC3H5O2, 1.3 x 10-5 Benzoic acid, HC7H5O2, 6.3 x 10-5 Which has the strongest conjugate base? A) propionic acid B) benzoic acid C) lactic acid D) formic acid

A

Rank acetic acid (HC2H3O2), hydrocyanic acid (HOCN), and hydrofluoric acid (HF) in order of increasing strength. Acid Ka HC2H3O2 1.8 x 10-5 HOCN 3.5 x10-4 HF 6.8 x 10-4 A) HC2H3O2 < HOCN < HF B) HOCN < HC2H3O2 < HF C) HF < HOCN < HC2H3O2 D) HF < HC2H3O2 < HOCN

A

You have 100.0 mL of 0.100 M aqueous solutions of each of the following acids: HCN, HF, HCl, and HC2H3O2. You titrate each with 0.100 M NaOH (aq). Rank the pHs of each of the solutions when each are titrated to the equivalence point, from highest to lowest pH. Ka for HCN = 6.2 10-10 Ka for HF = 7.2 10-4 Ka for HC2H3O2 = 1.8 10-5 A) HCN, HC2H3O2, HF, HCl B) HCl, HF, HCN, HC2H3O2 C) HF, HCN, HC2H3O2, HCl D) HC2H3O2, HCl, HCN, HF

A

H2PO4⁻+ HBO32- ---> HPO42- +H2BO3- The equilibrium constant for the reaction represented by the equation above is greater than 1.0. Which of the following gives the correct relative strengths of the acids and bases in the reaction? Acids Bases A) H2PO4- > H2BO3 - and HBO32- > HPO42- B) H2BO3- > H2PO4- and HBO32- > HPO42- C) H2PO4- > H2BO3- and HPO42- > HBO32- D) H2BO3- > H2PO4- and HPO42- > HBO32-

ANS: A If K is greater than 1.0, then products are favored and in the acid base world, that means they do not dissociate as much and are weaker than the conjugate on the left of the equation.

Commercial vinegar was titrated with NaOH solution to determine the content of acetic acid, HC2H3O2. For 20.0 milliliters of the vinegar, 32.0 milliliters of 0.500-molar NaOH solution was required. What was the concentration of acetic acid in the vinegar if no other acid was present? A) 1.60 M B) 0.800 M C) 0.640 M D) 0.400 M

ANS: B Expect easy math! Since NaOH and the monoprotic acid HC2H3O2 are involved you can use the shortcut formula for neutralization and solve for Ma. MaVa=MbVb Ma = MbVb/Va =32 x 0.5/20 = 16/20 = 8/10 = 0.8M

All of the following species can function as Brönsted-Lowery bases in solution EXCEPT A) H2O B) NH3 C) NH4⁺ D) HCO3⁻

ANS: C Brönsted-Lowery bases accept a proton (H+). The following can be formed H3O+, NH4 +, and H2CO3, but NH5 2+, does NOT exist.

2 NH3 ⇌ NH4+ + NH2- In liquid ammonia, the reaction represented above occurs. In the reaction NH4 + acts as A) a catalyst B) both an acid and a base C) the conjugate acid of NH3 D) the conjugate base of NH3

ANS: C Think of this reaction as you do the autoionization of water. Ammonia is definitely not a catalyst as well as the only reactant present. The ammonium ion accepted a proton (H+) from ammonia and since it is a reversible reaction, can donate the proton in which case it behaves as an acid.

Which of the following solutions has a pH greater than 7.0? A) 0.10 M KBr B) 0.10 M NH4Cl C) 0.10 M HC2H3O2 D) 0.10 M NaF

ANS: D A "pH greater than 7.0" is code for "basic or alkaline". You are now in search of a base or a basic salt. So, aceitic acid, HC2H3O2, is out right away! That leaves 3 salts. For each ask yourself which acid reacted with which base to form each salt and then is that acid or base strong or weak. Strong wins and drives the pH of the salt in solution. KBr--KOH (strong) + HBr (strong), therefore no victor, thus a neutral salt with a pH = 7.0 NH4Cl--NH4OH (weak) + HCl (strong), therefore acidic salt with a pH < 7.0 NaF--NaOH (strong) + HF (weak), therefore basic salt with a pH > 7.0

the PH of 0.1 molar sodium hydroxide is A) 1 B) 4 C) 11 D) 13

ANS: D Expect easy math! Realize that sodium hydroxide is a strong BASE! So, it's easiest to calculate pOH and then subtract from 14. pOH = -log [OH-] = -log [10-1], so, the log of [10-1] is simply -1, so the negative of -1 is plain old 1.0 [reported to 1 sig. fig.*]. Therefore the pH = 14 - 1 = 13.0 [still reported to 1 sig. fig.] * Recall that sig. fig. rules for pH are different since it is a logarithmic scale. The number in front of the decimal (the characteristic) is just a place holder, so the only significant figures are those after the decimal (the mantissa). It's explained in the appendix of your text if you don't believe me!

Which of the following reactions does NOT proceed significantly to the right in aqueous solutions? A) HCN + OH- → H2O + CN- B) Cu(H2O)4²⁺ + 4 NH3 →Cu(NH3)4²⁺ + 4H2O C) H2SO4 + H2O → H3O+ + HSO4⁻ D) H2O + HSO4⁻ → H2SO4 + OH-

ANS: D Sulfuric acid is a strong acid meaning it dissociates completely in water. This reaction represents the opposite of that and is not at all likely!

What is the pH of a 1.0 ́ 10-2 M solution of HCN? (For HCN, Ka = 4.0 ́ 10-10.) A) 10 B) Between 7 and 10 C) 7 D) Between 4 and 7

ANS:D HCN<-> H+ + CN- Ka=[H=][CN-]/[HCN] x^2/0.01=4x10^-10 x=2x10^-6 pH=-log2x10-6

If 25.0 mL of 0.451 M NaOH solution is titrated with 0.253 M H2SO4, the flask at the endpoint will contain (besides the indicator phenolphthalein) as the principal components: A) sodium hydroxide, sulfuric acid, and water B) dissolved sodium sulfate and water C) sodium hydroxide, sodium sulfate, and water D) dissolved sodium sulfate, sulfuric acid, and water

B

In titrating 0.20 M hydrochloric acid, HCl, with 0.20 M NaOH at 25°C, the solution at the equivalence point is A) 0.20 M NaCl B) 0.10 M NaCl C) slightly acidic D) 0.10 M HCl and 0.20 M NaOH

B

Consider the following indicators and their pH ranges: Methyl orange 3.2-4.4 Methyl red 4.8-6.0 Bromothymol blue 6.0-7.6 Phenolphthalein 8.2-10.0 Alizarin yellow 10.1-12.0 Assume an indicator works best when the equivalence point of a titration comes in the middle of the indicator range. For which of the following titrations would methyl red be the best indicator? A) 0.100 M HNO3 + 0.100 M KOH B) 0.100 M aniline (Kb = 3.8 10-10) + 0.100 M HCl C) 0.100 M NH3 (Kb = 1.8 10-5) + 0.100 M HCl D) 0.100 M HF (Ka = 7.2 10-4) + 0.100 M NaOH

C

For the equilibrium that exists in an aqueous solution of nitrous acid (HNO2, a weak acid), the equilibrium-constant expression is

C

In deciding which of two acids is the stronger, one must know: A) the concentration of each acid solution B) the pH of each acid solution C) the equilibrium constant of each acid D) both A and C must be known

C

As water is heated, its pH decreases. This means that: A) The water is no longer neutral. B) The Kw value is decreasing. C) The water has a lower [OH-] than cooler water. D) The dissociation of water is an endothermic process.

D

You have 100.0 mL of a solution of hydrochloric acid that has a pH of 3.00. You add 100.0 mL of water to this solution. What is the resulting pH of the solution? A) The pH = 5.00 (the average of 3.00 and 7.00). B) The pH = 10.00 (3.00 + 7.00 = 10.00). C) The pH = 3.00 (water is neutral and does not affect the pH). D) None of the above is correct, but the pH must be greater than 3.00.

D

its a picture with a beaker with a pH of 7.00 and has cations and anions Which of the following salts is most likely to form an aqueous solution having the pH shown in the figure above? A) Na2CO3 B) NH4Cl C) Zn(NO3)2 D) KCl

D Remember, acid + base yields salt + water. So, work backwards! Which acid reacted with which base to form the salt in question AND remember, strong wins! A) strong base + weak acid forms basic salt. B) & C)weak base + strong acid forms acidic salt. D) strong acid + strong base forms neutral salt.

Which of the following would be the best mole ratio of conjugate acid to conjugate base for a large buffer capacity? A) 100:1 B) 1:1 C) 1:100 D) 1:10

Designing buffers for specific environments takes the A:B ratio into account. A) has a large amount of weak acid (100) versus weak base (1). But what does that mean...it means the buffer can neutralize large amounts of base invaders, but next to no acid invaders. Vice versa for C). D) can neutralize 10 times more invading base than acid. B) has the largest capacity (1:1) indicating that this buffer can "neutralize" either an acidic or basic invader equally well with regard to quantity.