CHEM 1332 - Ch. 13

Pataasin ang iyong marka sa homework at exams ngayon gamit ang Quizwiz!

Use a table of Ka or Kb values to determine whether K c for the following reaction is less than, equal to or greater than 1 NH4+ + HCO3- <=> H2CO3 + NH3 a) Kc > 1 b) Kc = 1 c) Kc < 1

C

According to the VSEPR model, the electron-pair arrangement of the central atom in BH3 is predicted to be __________. a) linear b) trigonal planar c) tetrahedral d) trigonal bipyramidal c) tetrahedral

trigonal planar

Classify H2Se a) strong acid b) weak acid c) strong base d) weak base

weak acid

Classify H2SeO4 a) strong acid b) weak acid c) strong base d) weak base

weak acid

Classify NH3 a) strong acid b) weak acid c) strong base d) weak base

weak base

Is an aqueous solution of NaNO2 acidic, basic or neutral (use A, B or N)?

B

Which of the following acids has the strongest conjugate base? a) 2-hydroxybenzoic acid, Ka = 1.1 x 10-3 b) Benzoic acid, Ka = 6.5 x 10-5 c) 3-chlorobenzoic acid, Ka = 1.5 x 10-4 d) Chloroacetic acid, Ka = 1.4 x 10-3 e) Ascorbic acid, Ka = 8.0 x 10-5

B

Is an aqueous solution of LiNO2 acidic, basic or neutral (use A, B or N)?

Basic

The Ka of hypochlorous acid (HClO) is 3.0 × 10-8 at 25.0 °C. Calculate the pH of a 0.0345 M hypochlorous acid solution.

4.49

What is the pH of a room temperature solution with a concentration of OH- of 9.69 x 10-10? Remember: Kw = 10-14, and pH + pOH = 14= pKw

4.986 ± 0.01 Response Feedback: [H+] = 10⁻¹⁴/[OH-] and then pH = -log[H+] or -log[OH-] = pOH and then pH = 14 - pOH

What is the pOH of a 0.435 M solution of CsOH? Enter your answer with two decimal places.

0.36 ± 0.02

Match each substance with the correct designation for the equation HSO3- + CH3NH2 <=> SO32- + CH3NH3+ 1. HSO3- 2. CH3NH2 3. SO32- 4. CH3NH3+ A. conjugate base of HSO3- B. conjugate acid of SO32- C. conjugate acid of CH3NH2 D. conjugate base of CH3NH3+

1. B 2. D 3. A 4. C

What is the pOH of a 0.091 M solution of CsOH? Enter your answer with two decimal places.

1.04 ± 0.02

What is the pOH of a 0.045 M solution of calcium hydroxide? Hint: remember to write the chemical formula and the consider the stoichiometry.

1.05

What is the pOH of a 0.030 M solution of calcium hydroxide? Hint: remember to write the chemical formula and the consider the stoichiometry.

1.22

The Ka of hydrofluoric acid (HF) at 25.0 °C is 6.8 × 10-4. What is the pH of a 0.45 M aqueous solution of HF?

1.76

Calculate the pH of a 0.288 M solution of a base that has a Kb = 3.98 x 10-7.

10.53 ± 0.02 Response Feedback: For base problems, remember to first calculate the hydroxide concentration from the base ionization equilibrium and the Kb value: B + H2O ⇌ HB+ + OH-. Once you know the hydroxide concentration, you can calculate pOH and then pH = 14 - pOH.

Calculate the pH of a 0.10 M solution of ammonia (NH3, Kb = 1.8 x 10-5.) a) 5.13 b) 9.26 c) 11.13 d) 8.87 e) 2.87

11.13

What is the pH of a room temperature solution with a concentration of OH- of 1.35 x 10-2? Remember: Kw = 10-14, and pH + pOH = 14= pKw

12.130 ± 0.01 Response Feedback: [H+] = 10⁻¹⁴/[OH-] and then pH = -log[H+] or -log[OH-] = pOH and then pH = 14 - pOH

Calculate the pH of a 0.60 M solution of ethylamine (C2H5NH2, Kb = 5.6 x 10-4.) a) 5.49 b) 8.51 c) 1.74 d) 10.75 e) 12.26

12.26

What is the pH of a 0.020 M aqueous solution of the soluble salt, barium hydroxide? Hint: remember to write the chemical formula and the consider the stoichiometry.

12.60

What is the pH of a room temperature solution with a concentration of OH- of 4.35 x 10-1? Remember: Kw = 10-14, and pH + pOH = 14= pKw

13.638 ± 0.01 Response Feedback: [H+] = 10⁻¹⁴/[OH-] and then pH = -log[H+] or -log[OH-] = pOH and then pH = 14 - pOH

Calculate the pH of a 0.300 M solution of benzoic acid, for which the Ka value is 6.50 x 10-5. a) 2.35 b) 4.42 x 10-3 c) 11.65 d) 5.17 e) 8.83

2.35

What is the concentration of hydroxide ion in an aqueous solution at 25.0 °C that has a pOH of 3.55?

2.8 × 10-4 Response Feedback: pH = - log[H+] which means [H+] = 10⁻ᵖᴴ. pOH = - log[OH-] which means [ OH- ] = 10⁻ᵖᴼᴴ.

If the pH of an aqueous solution at 25.0 °C is 10.55, what is the molarity of H+ in this solution?

2.8 × 10⁻¹¹ Response Feedback: pH = - log[H+] which means [H+] = 10⁻ᵖᴴ. pOH = - log[OH-] which means [ OH- ] = 10⁻ᵖᴼᴴ.

What is the pH of a 0.0015 M solution of HClO4? a) 5.65 b) 0.0015 c) 11.18 d) 2.82

2.82

Calculate the pH of a 0.200 M solution of carbonic acid, for which the Ka value is 4.50 x 10-7. a) 10.48 b) 4.18 c) 3.00 x 10-4 d) 9.82 e) 3.52

3.52

Calculate the pH of a 0.200 M solution of hypochlorous acid, for which the Ka value is 2.90 x 10-8. a) 9.88 b) 3.58 c) 7.62 x 10-5 d) 10.42 e) 4.12

4.12

At a certain temperature an aqueous solution of 0.0604 M monoprotic acid (HA) is measured to be 0.0833 % dissociated (e.g. ionized). What is the Ka value for this acid?

4.19E-8 ± 2% Response Feedback: The concentration of the ions can be calculated by using the percentage and initial concentration to calculate the concentration of ions formed. You can then use your calculated equilibrium concentrations to substitute into the equilibrium expression and obtain the value of Ka.

Calculate the pOH of a 0.0987 M aqueous sodium cyanide solution at 25.0 °C. Kb for CN- is 4.9 x 10-10.

5.16 Response Feedback: Remember, in the equilibrium reaction for Kb the base reacts with water to form hydroxide: B + H2O ⇌ HB + OH- Since you are given the value of Kb, you can solve for [OH-] and then calculate pOH = - log[OH-] Keep in mind that pH could easily be calculated using pH = 14 - pOH if needed.

What is the pH of 0.77 M methylammonium bromide, CH3NH3Br? Enter your answer with two decimal places. The Kb of CH3NH2 is 4.2 x 10-4.

5.37 ± 0.02

What is the pH of 0.65 M triethylammonium iodide, (C2H5)3NHI. Enter your answer with two decimal places.The Kb of triethylamine, (C2H5)3N, is 5.2 x 10-4.

5.45 ± 0.02

What is the pH of 0.395 M dimethylammonium iodide, (CH3)2NH2I? Enter your answer with two decimal places.The Kb of (CH3)2NH is 5.9 x 10-4.

5.59 ± 0.02

What is the pH of a room temperature solution with a concentration of OH- of 7.14 x 10-7? Remember: Kw = 10-14, and pH + pOH = 14= pKw

7.854 ± 0.01 Response Feedback: [H+] = 10⁻¹⁴/[OH-] and then pH = -log[H+] or -log[OH-] = pOH and then pH = 14 - pOH

Calculate the pOH of 0.486 M anilinium hydrochloride (C6H5NH3Cl) solution in water, given that Kb for aniline is 3.83 × 10-4. This one has several steps so here are a few hints: 1) anilinium hydrochloride (C6H5NH3Cl) is the conjugate acid of aniline (C6H5NH2) 2) The Ka value for anilinium hydrochloride (C6H5NH3Cl) can be found using the relationship Kw = Ka*Kb 3) If you set-up the ICE-table for the reaction: C6H5NH3+ ⇌ H+ + C6H5NH2 and solve for [H+], you can then calculate pH and finally pOH.

8.55 Response Feedback: Step check points: the Ka = 2.61E-11 since Kb=Kw/Ka the [H+] = 3.56E-6 the pH = 5.45 (since 14 = pH + pOH) the pOH = 8.55

A 0.411 M aqueous solution of weak monoprotic acid (HA) has a pH of 4.504, what is the pKa of the acid dissociation constant?

8.62 ± 0.02 Response Feedback: The pH can be used to calculate the equilibrium concentration of ions which can then be substituted into the equilibrium expression to obtain the value of Ka. The pKa is then calculated as -log(Ka).

Calculate the pH of a 0.40 M solution of sodium formate (NaHCOO) given that the Ka of formic acid (HCOOH) is 1.8 x 10-4. a) 11.93 b) 2.07 c) 5.33 d) 10.26 e) 8.67

8.67

Calculate the pH of a 0.20 M solution of sodium benzoate (NaC6H5COO) given that the Ka of benzoic acid (C6H5COOH) is 6.50 x 10-5. a) 2.44 b) 9.81 c) 5.26 d) 11.56 e) 8.74

8.74

Calculate the pH of a 0.50 M solution of sodium benzoate (NaC6H5COO) given that the Ka of benzoic acid (C6H5COOH) is 6.50 x 10-5. a) 2.24 b) 9.81 c) 11.76 d) 8.94 e) 5.06

8.94

Calculate the pH of a 0.60 M solution of aniline(C6H5NH2, Kb = 3.8 x 10-10.) a) 9.18 b) 2.40 c) 4.82 d) 9.42 e) 11.60

9.18

Consider the reaction shown. Using your knowledge of relative acid-base strengths and equilibrium, determine what you can about the size of Kc for the reaction. HOCl(aq) + ClO3-(aq) <=> OCl-(aq) + HClO3(aq) a) K c < 1 b) K c > 1 c) K c = 1 d) Impossible to tell

A

Which of the following acids has the strongest conjugate base? a) 2-hydroxybenzoic acid, Ka = 1.1 x 10-3 b) Chlorous acid, Ka = 1.1 x 10-2 c) Chloroacetic acid, Ka = 1.4 x 10-3 d) Paraperiodic acid, Ka = 12.8 x 10-2 e) Iodic acid, Ka = 1.7 x 10-1

A

What is the conjugate base of HClO4 ?

ClO4-

Which of the following is not a strong acid in aqueous solution? a) HCl b) HI c) HClO4 d) HBr e) H2SO3

H2SO3

Of the following, which is the weakest acid? a) HClO4 b) HClO3 c) HClO2 d) HClO e) HCl

HClO Response Feedback: For oxyacids, for a given central atom, the acidity increases with increasing number of oxygen atoms.

Which of the following is not a strong acid in aqueous solution? a) HNO3 b) HI c) H2SO4 d) HClO2 e) HBr

HClO2

Which of the following is not a strong acid in aqueous solution? a) HNO2 b) HI c) H2SO4 d) HBr e) HClO4

HNO2

Select the stronger acid from each of the following pairs: I. HI or HBr II. H3AsO3 or H2SeO3 III. HNO3 or HNO2

I. HI II. H3AsO3 III. HNO3

Consider the reaction shown. Using your knowledge of relative acid-base strengths and equilibrium, determine what you can about the size of K c for the reaction. H3AsO4(aq) + HSeO4-(aq) <=> H2AsO4-(aq) + H2SeO4(aq)

Kc < 1

Use a table of Ka or Kb values to determine whether K c for the following reaction is less than, equal to or greater than 1 NH4+ + HCO3- <=> H2CO3 + NH3

Kc < 1

A 0.1 M aqueous solution of ________ will have a pH of 7.0 at 25 oC. a) LiNO2 b) LiBr c) Ba(ClO4)2 d) NH4ClO4

LiBr and Ba(ClO4)2 Response Feedback: Conjugate bases of strong acids are neutral (i.e. not bases). (Based on the list of seven strong acids, this mean the ions HSO4-, NO3-, I-, Br-, Cl-, ClO3-, and ClO4- can form neutral salts. Group I and II cations form neutral salts, but since ammonium is the conjugate base of a weak acid (ammonia) it forms an acidic salt. Anions that are conjugate bases of weak acids form basic salts.

A 0.1 M aqueous solution of ________ will have a pH of 7.0 at 25 oC. a) LiF b) LiBr c) NaClO4 d) NH4I

LiBr and NaClO4 Response Feedback: Conjugate bases of strong acids are neutral (i.e. not bases). (Based on the list of seven strong acids, this mean the ions HSO4-, NO3-, I-, Br-, Cl-, ClO3-, and ClO4- can form neutral salts. Group I and II cations form neutral salts, but since ammonium is the conjugate base of a weak acid (ammonia) it forms an acidic salt.

What is the conjugate base of NH4+ ?

NH3

Which of the following acids has the strongest conjugate base? a) H2Te b) H2O c) H2Se d) H2S e) NH3

NH3

Is an aqueous solution of LiBr acidic, basic or neutral (use A, B or N)?

Neutral

Give the conjugate base of HPO32-. Write it ignoring super and subscripts (so, HPO42- would be HPO42-)

PO33-

What is the conjugate base of HSO4- ?

SO42-


Kaugnay na mga set ng pag-aaral

Unit 5 Quiz and Test Questions, Exam 5 Microbiology ALL

View Set

Chapter 33: Assessment and Management of Patients with Allergic Disorders

View Set

Pharmacology NCLEX style questions

View Set