Question

Please explain these:

34. Which solution has the highest pH?

Ka:

HCN 5.8×10–10

CH3COOH 1.8×10–5

(A) 0.10 M CH3COOH (B) 0.10 M HCN (C) 0.10 M CH3COOK (D) 0.10 M NaBr

7. A dilute solution of which acid is most likely to produce a reduction product other than H2 when it reacts with a metal? (A) HF (B) HCl (C) HNO3 (D) H2SO4

35. A 75 mL solution that is 0.10 M in HC2H3O2 and 0.10 M in NaC2H3O2 has a pH of 4.74. Which of the following actions will change the pH of this solution?

I Adding 15 mL of 0.10 M HCl

II Adding 0.010 mol of NaC2H3O2

III Diluting the solution from 75 mL to 125 mL

(A) I only (B) II only (C) I and II only (D) I, II and III

Solutions:

34) C

7) C

35) C

Answer 1

7 ) c) It also can produce oxides of nitrogen.

34) C)  assuming that initial concentration of each is 0.1 M

for CH3COOH
1.8 x 10^-5 = x^2 / 0.1-x
x = [H+]= 0.0013 M
pH = 2.9

for HCN
5.8 x 10^-10 = x^2 / 0.1-x
x = [H+]= 7.6 x 10^-6 M
pH = 5.1

for CH3COO- Kb = Kw/Ka = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.6 x 10^-10
CH3COO- + H2O <=> CH3COOH + OH-
5.6 x 10^-10 = x^2 / 0.1-x
x = [OH-]= 7.5 x 10^-6 M
pOH = 5.1
pH = 14 - 5.1 = 8.9

NaBr is a neural salt : pH = 7

35) c) 10 M = 0.000951

total volume = 0.0845 L

CH3COOH + OH- >> CH3COO- + H2O

moles acetic acid = 0.0075 - 0.000951 =0.00655
[CH3COOH]= 0.00655/ 0.0845 L =0.0775 M
moles acetate = 0.015 + 0.000951 =0.0160
[CH3COO-]= 0.0160 /0.0845 L =0.189 M