Exam 2A, Spring 2018

1. Complete and balance the following in aqueous solution.

a. cyanate ion plus perchloric acid

b. hydrogen phosphate ion plus magnesium hydroxide

c. hydroiodic acid plus rubidium hydroxide

Answer

a. OCN–(aq) + HClO4(aq) → HOCN(aq) + ClO4–(aq)

b. 2 HPO42–(aq) + Mg(OH)2(aq) → 2 H2O(l) + Mg2+(aq) + 2 PO43–(aq)

c. HI(aq) + RbOH(aq) → H2O(l) + Rb+(aq) + I–(aq)

2. Write the mass action expression for Kc for each of the following:

a. P4(s) + H2(g) → ← PH3(g)

b. CH4(g) + Cl2(g) → ← CCl4(l) + HCl(g)

c. Fe(s) + CO(g) → ← Fe(CO)5(l)

Answer

a. Balance: P4(s) + 6 H2(g) → ← 4 PH3(g)

Kc = [PH3]e4              [H2]e6

b. Balance: CH4(g) + 4 Cl2(g) → ← CCl4(l) + 4 HCl(g)

Kc = [HCl]e4                          [CH4]e[Cl2]e4

c. Balance: Fe(s) + 5 CO(g) → ← Fe(CO)5(l)

Kc = 1              [CO]e5

3. Consider the reaction 2 NaHCO3(s) → ← Na2CO3(s) + H2O(g) + CO2(2). Which does the reaction shift if:

a. the humidity rises?

b. sodium carbonate is added to the reaction?

Answer

a. Rising humidity increases the concentration of water vapor, which will shift the reaction towards reactants, to the left.

b. sodium carbonate is a solid so does not affect the equilibrium so there is no shift.

4. Write the appropriate balanced reaction using Brønsted-Lowry theory, find the hydronium ion concentration, the hydroxide ion concentration, pH, and pOH for the following, including units:

a. A 0.010 M solution of hydrochloric acid

b. A 0.010 M solution of potassium hydroxide

Answer

a. HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)

[H3O+] = [HCl] = 0.010 M

[OH–] = Kw/[H3O+] = 1.0×10–14/0.010 = 1.0×10–12 M

pH = –log[H3O+] = –log(1.0×10–2) = 2.00

pOH = 14.00 – pH = 14.00 – 2.00 = 12.00

b. KOH(aq) → K+(aq) + OH–(aq)

[OH–] = [KOH] = 0.010 M

[H3O+] = Kw/[OH–] = 1.0×10–14/0.010 = 1.0×10–12 M

pOH = –log[OH–] = –log(1.0×10–) = 2.00

pH = 14.00 – pOH = 14.00 – 2.00 = 12.00

5. Find the pH and per cent ionization for a 0.0010 M solution of hypoiodous acid, HIO. Show all of your work. Ka = 2.3×10–11

Answer

HIO(aq)+ H2O(l) → ← H3O+(aq)+ IO–(aq)

Ka = [H3O+]e[IO–]e                        [HIO]e = 2.3×10–11

Initial0.001000

Change–x+x+x

Equilibrium0.0010 – xxx

Approximate? 0.0010/2.3×10–11 = 4.3×107 > 100 so Yes

2.3×10–11 = x2/0.0010

x = [H3O+]e = 1.5×10–7 M

pH = –log[H3O+] = –log(1.5×10–7) = 6.82

α = ([H3O+]e/[HIO]init)×100% = (1.5×10–7)(100)/(0.0010) = 0.015 %