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1. Which of the following is NOT a valid definition of an acid?(10 points) (A) A

ID: 993000 • Letter: 1

Question

1. Which of the following is NOT a valid definition of an acid?(10 points)

(A) An electron pair acceptor

(B) A substance that lowers the pOH of a solution

(C) A proton donor

(D) A substance that produces H+ in aqueous solution

2. If the Ka of sulfurous acid is 1.54 x 10-2, what is the Kb of its conjugate base? (10 points)

(A) 1.54 x 10-12

(B) 6.49 x 10-13

(C) 1.30 x 10-12

(D) 6.17 x 10-10

3. What is the Ka of a 0.320 M solution of nitrous acid? Nitrous acid has a pKa of 3.34. (15 points)

(A) 0.320

(B) 2.29 x 10-4

(C) 4.57 x 10-4

(D) 4.57 x 10-6

4. Which of the following is true? (15 points)

(A) pH + pK a = 14

(B) pKb + pOH = 14

(C) pH + pOH = Kw

(D) pKa + pKb = pKw

5. During a titration the following data were collected. A 20.0 mL portion of solution of an unknown acid HX was titrated with 2.0 M KOH. It required 60.0 mL of the base to neutralize the sample. What is the molarity of the acid HX? (15 points)

                A)            6.0 M

                B)            0.67 M

                C)            3.0 M

                D)            0.12 M

6. An acid-base titration curve plots(15 points)

                A)            the pH of a solution against the total volume of an added acid

                B)            the pH of a solution against the volume of an added acid or base

                C)            the pH of a solution over time

                D)            the pH of a solution containing buffers

Explanation / Answer

1) Answer is A. Acid is not an electron pair acceptor, it is donor.

2) For a conjugate acid-base pair, Ka × Kb = Kw where Kw is 1x10-14

Then Kb = Kw/Ka =  1x10-14 / 1.54 x 10-2 = 6.49 x 10-13

3) pKa = -log Ka

3.34 = -log Ka, then Ka = antilog(-3.34)= 4.57x 10-4

4) answer is D. pKa+pKb = pKw

5) No. of moles of KOH required to neutralise the acid = 2 M/1000mL * 60 mL = 0 .12 mols

So 20 mL of acid contains 0.12 mols of acid. Then molarity of the acid = 0.12 mols / 20 mL * 1000 mL = 6M

6) answer is B