Exam 2B, Spring 2018

1. Complete and balance the following in aqueous solution.

a. thiocyanate ion plus perchloric acid

b. hydrogen phosphate ion plus nitric acid

c. hydrobromic acid plus cesium hydroxide

Answer

a. SCN–(aq) + HClO4(aq) → HSCN(aq) + ClO4–(aq)

b. HPO42–(aq) + 2 HNO3(aq) → H3PO4(aq) + 2 NO3–(aq)

c. HBr(aq) + CsOH(aq) → H2O(l) + Cs+(aq) + Br–(aq)

2. Write the mass action expression for Kp 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)

Kp = PPH3e4              PH2e6

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

Kp = PHCle4                          PCH4ePCl2e4

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

KP = 1              PCOe5

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

a. the humidity decreases?

b. sodium bicarbonate is added to the reaction?

Answer

a. Decreasing humidity lowers the concentration of water vapor, which will shift the reaction towards products, to the right.

b. sodium bicarbonate 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 0.0010 M solution of hydrochloric acid

b. A 0.0010 M solution of lithium hydroxide

Answer

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

[H3O+] = [HCl] = 0.0010 M

[OH–] = Kw/[H3O+] = 1.0×10–14/0.0010 = 1.0×10–11 M

pH = –log[H3O+] = –log(1.0×10–3) = 3.00

pOH = 14.00 – pH = 14.00 – 3.00 = 11.00

b. LiOH(aq) → Li+(aq) + OH–(aq)

[OH–] = [LiOH] = 0.0010 M

[H3O+] = Kw/[OH–] = 1.0×10–14/0.0010 = 1.0×10–11 M

pOH = –log[OH–] = –log(1.0×10–3) = 3.00

pH = 14.00 – pOH = 14.00 – 3.00 = 11.00

5. Find the pH and per cent ionization for a 0.0010 M solution of hypobromous acid, HBrO. Show all of your work. Ka = 2.5×10–9

Answer

HBrO(aq)+ H2O(l) → ← H3O+(aq)+ BrO–(aq)

Ka = [H3O+]e[BrO–]e                        [HBrO]e = 2.5×10–9

Initial0.001000

Change–x+x+x

Equilibrium0.0010 – xxx

Approximate? 0.0010/2.5×10–9 = 4.0×105 > 100 so Yes

2.5×10–9 = x2/0.0010

x = [H3O+]e = 1.6×10–6 M

pH = –log[H3O+] = –log(1.6×10–6) = 5.80

α = ([H3O+]e/[HBrO]init)×100% = (1.6×10–6)(100)/(0.0010) = 0.16 %