CHEM 104 – Module 2 Exam – Portage Learning – Verified Questions and Answers for Exam Succ, Exams of Nursing

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CHEM 104 Module 2 Exam Portage Learning

Questions and Verified Answers, 100% Pass

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Portage Learning CHEM 104 Module 2 Exam

Question 1 Identify each of the compounds below as ACID, BASE or SALT on the basis of their formula andexplain your answer. (1) Cr(OH) 3 (2) HAsO 4 (3) CoCO 3 Your Answer:

1. base - contains Cr metal + OH polyatomic group
2. acid - contains H + polyatomic group (AsO 4 )
3. salt - contains Co metal + polyatomic group (CO 3 ) Question 2 For the Brønsted-Lowry acid base reactions shown below, list the stronger acid, stronger base, weaker acid and weaker base in the answer blanks provided: NH +^ + H PO -^ NH + H PO Stronger acid: Stronger base: Weaker acid: Weaker base: Your Answer: Stronger acid: H 3 PO 4 Stronger base: NH 3 Weaker acid: NH + 4 Weaker base: H 2 PO -

2 3 2 Question 5 In the titration of 15.0 mL of H 2 SO 4 of unknown concentration, the phenolphthalein indicator present in the colorless solution turns pink when 26.4 mL of 0.130 M NaOH is added. Show the calculation ofthe molarity of the H 2 SO 4. Your Answer: (Ma x mLa)/1000 x Sa/Sb = (Mb x mLb)/ (Ma x 15.0)/1000 x 1/2 = (0.130 x 26.4)/ MH2SO4 = 0.458 M ( H 2 SO 4 is the acid and Na OH is the base) Ma x mLa / 1000 x Sa / Sb = Mb x mLb / 1000 Max 15.0 / 1000 x 2 / 1 = 0.130 x 26.4 / 1000 MH2SO = 0.114 M ratio: 2 / 1 Question 6 Show the calculation of the [H+], pH and % ionization for 0.645 M acetic acid (HC 2 H 3 O 2 )HC H O H+^ + C H O -^ K = 1.8 x 10 -^5 2 3 2 a Your Answer: HC 2 H 3 O 2 H+^ + C 2 H 3 O 2 - 0.645 0 0

• x +x +x

0.645-x x x

2 3 2 2 3 2 4 pH = - log [H+] = - log (3.41x10-^3 )pH = 2. % ionization = ([C H O - ]/ [HC H O ]) x 100 % ionization = (3.41x10-^3 / 0.645) x 100 % ionization = 0.53% Question 7 Predict and explain whether a solution of LiF is acidic, basic or neutral. Your Answer: Basic since Li hydrolyzes to form strong base [LiOH]. LiF: Basic since F-^ hydrolyzes to form a weak acid (HF) and OH- Question 8 Show calculation of the pH of a buffer prepared by mixing 0.100 M NH 4 Cl and 0.0750 M NH 3. NH 3 + H 2 O (^) (liq) NH +^ + OH-^ Ka = 1.8 x 10 -^5 Your Answer:

NH 3 + H 2 O (^) (liq) NH 4 +^ + OH- 0.0750 0.100 0

• x +x +x

2 3 2 a 2 3 2 1.8 x 10-^5 = (0.100) (x) / (0.0750)x = 1.35x10-^5 = [OH-] pOH = - log [OH-] = - log (1.35x10-^5 ) pOH = 4. pH = 14 - pOH = 14 - 4. pH = 9. Question 9 Show the calculation of the pH of a solution obtained by adding 0.0100 mole of OH-^ to a buffer of pH 4.74 which originally contains 0.100 M NaC 2 H 3 O 2 and 0.100 M HC 2 H 3 O 2. HC 2 H 3 O 2 H+^ + C H O -^ K = 1.8 x 10 -^5 Your Answer: HC 2 H 3 O 2 H+^ + C H O - 0.100 0 0. +0.0100 mole OH- 0.090 0 0.

• x +x +x 0.090-x Ka = 1.8 x 10 -^5 x 0.110+x Ka = [H+] [C 2 H 3 O 2 - ] / [HC 2 H 3 O 2 ] 1.8 x 10 -^5 = (x) (0.110+x) / (0.090-x ) 1.8 x 10 -^5 = (x) (0.110) / (0.090)x = 1.47x10-^5 = [H+] pH = - log [H+] = - log (1.47x10-^5 )pH = 4.

pH went up slightly fron 4.74 to 4.83 when OH-^ was added.