FPC May 23 workshop questions

Question 1

Calculate the pH of the following solution

0.05 M Mg(OH)2

Answer 1

[OH-] = 2 x [Mg(OH)2] = 0.10 mol dm3
pOH = -log(0.10) = 1
pH = 13.0

Question 2

Identify the conjugate pairs in the following equation indicating the Bronsted acid and base in each pair
HF + CN- <=> HCN + F-

Answer 2

Acid + Conjugate base <=> Base + conjugate acid
HF + CN- <=> HCN + F-

Question 3

Calculate the pH of a solution being 0.50 mol dm-3 in sodium methanolate and 0.30 mol dm-3 in methanoic acid (Ka = 1.78 x 10*-4)

Answer 3

Pka = -log (1.78 x 10*-4) = 3.75

pH = 3.75 + log(0.50/0.30) = 3.97

Question 4

Calculate the pH at the start of a titration of 25cm3 of a 0.01 moldm-3 solution of ethanoic acid (pKa = 4.75) with NaOH (0.003 mol dm-3) and then after 5cm of base has been added

Answer 4

H+ at the start = Sqrt Ka x c = Sqrt (1.78x10-5) x c = 4.22 x 10-4

pH = -log (4.22 x 10*-4 ) = 3.38

After 5cm

pH = pKa + log(salt/acid0

pH = pKa + log(5 x 0.003) / ((25 x 0.01) - (5 x 0.003)) = pKa + log (0.015/0.235) = 3.56

Question 5

The resistance of a conductance cell containing 0.10 mol dm-3 lithium nitrate solution is 192 Ω (ohm). The resistance of the same cell containing 0.02 mol dm- 3 potassium chloride solution is 620 is Ω . The standard conductivity of 0.02 mol dm-3 potassium chloride is 0.242 S m-1.

Calculate

i) cell constant of conductance cell
ii) conductivity of the lithium nitrate solution
iii) Molar conductivity of lithium nitrate solution

Answer 5

i) 150 m-1
ii)0.781 Sm-1
iii) 0.00781 Sm2 mol-1

Question 6

Calculate the molar conductviity at infinite dilution of butric acid, given the following molar conductivities at infinite dilution.

Sodium butyrate 0.00826 S m2 mol-1
Hydrochloric acid 0.0426 Sm2 mol-1
Sodium chloride 0.0126 S m2 mol-1

Answer 6

Λ ∞ = H But = Λ∞ (H+) + Λ∞ (But-) = Λ∞ (NaBut) + Λ∞(HCl) + Λ∞ (But-)

Λ∞(Na+) + Λ∞(But-) + Λ∞(H+) + Λ∞(Cl-) - Λ∞(Na+) - Λ∞(Cl-)

= Λ∞ (H But) = 0.00826 + 0.0426 - 0.0126 = 0.0383 Sm2mol-1

Question 7

The solubility of barium sulphate is 2.56 x 10*-6 kg dm-3. If the limiting molar conductivity of barium sulphate is 2.865 x 10-2 S m2 mol-1 and water has a conductivity of 1.05x10-3 S m-1, what is the conductivity of a saturated solution of the salt (Mr BaSO4 = 233)

Answer 7

Λ∞ = K/C

[BaSO4] = 2.56 x 10-6 kg dm-3 / 233 x 103 kg mol = 1.099 x 10-5

KBaSo4 = [BaSO4] x Λ∞ = (1.099 x 10-5) x (2.865 x 10-2) = 3.15 x 10*-4

(3.15 x 10-4) + (1.05X10-4) = 4.20 x 10*-4 Sm-1

Question 8

Calculate the volume occupied by one mole of a gas at 25 oC and 100 kPa.

Answer 8

𝑃𝑉 = 𝑛RT => 𝑉 = 𝑛𝑅𝑇 / P

25 degrees C = 298K
100kPa = 1.00 x 10*5 Pa
Pa = J m-3

1mol x 8.314 JK mol x 298K / 1.00 x 10*5 Jm-3= 0.0248m3 = 24.8dm3

Question 9

Calculate the temperature of a gas if 0.5 moles occupy 1.2 dm3 at a pressure of 200 kPa.

Answer 9

T = PV / nR

1.2 dm3 = 0.0012 m3
200 kPa = 2.00 x 105 Pa
Pa = J m-3

(2.00 x 10*5) x 0.0012 / 0.05 x 8.314 = 57.7

Question 10

Calculate the relative molecular mass of a gas if a 500 cm3 sample at 20C and 1 atm has a mass of 0.66 g

Answer 10

G/moles = Rmm

n = PV / RT

1.01325 x 10*5 Jm-3 x 0.0005 / 8.314 Jk x 293K

Question 11

A system has constant volume (ΔV=0) and the heat around the system increases by 45 J.

a. What is the sign for heat (q) for the system?
b. What is ΔU equal to?
c. What is the value of internal energy of the system in Joules?

Answer 11

Since the system has constant volume (ΔV=0) the term -PΔV=0 and work is equal to zero (w=0). Thus, in the equation ΔU= q + w, ΔU=q. The internal energy is equal to the heat of the system. The surrounding heat increases, so the heat of the system decreases, because heat is neither created nor destroyed. Therefore, heat is taken away from the system making it exothermic and negative. The value of Internal Energy will be the negative value of the heat absorbed by the surroundings.

a. negative (q<0)
b. ΔU=q + (-PΔV) = q+ 0 = q
c. ΔU = -45J

Question 12

If 1200 J of heat are added to a system and the system does 800J of work on the surroundings what is the internal energy change of the system

Answer 12

U = q + w = = +1200 J + (-800 J) = ΔU (system) = + 400 J

Question 13

What is the change in internal energy (U) of a system that absorbs 2500 J of heat and does 7655 J of work on the surroundings ?

Answer 13

ΔU = q + w = = +2500 J + (-7655) = -5155 J

Question 14

If the change in internal energy for the combustion of 1 mole of methane at 298 K is -885 kJ mol-1 , what is the enthalpy associated with this process?

Answer 14

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)
∆𝐻 = ∆𝑈 + ∆𝑛𝑅𝑇

∆𝑛 = 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝒈𝒂𝒔𝒆𝒐𝒖𝒔 𝑚𝑜𝑙𝑒𝑠 = 1 − 3 = −2

∆𝐻 = ∆𝑈 + ∆𝑛𝑅𝑇 = −885000 𝐽𝑚𝑜𝑙−1 + (−2) × 8.314 𝐽 𝐾 −1𝑚𝑜𝑙−1 × 298 𝐾
∆𝐻 = −889955 𝐽𝑚𝑜𝑙−1 = −890 𝑘𝐽 𝑚𝑜𝑙−1

𝑛𝑜𝑡𝑒 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 ∆𝐻 𝑎𝑛𝑑 ∆𝑈 𝑎𝑟𝑒 𝑠𝑖𝑚𝑖𝑙𝑎r

Question 15

Identify Heisenberg’s uncertainty

Answer 15

∆x x m x ∆p > h / 4pi

Question 16

Identify hess’ law

Answer 16

∆𝐻𝑛𝑒𝑡 = net change in enthalpy for a series of reactions
E∆𝐻𝑟 = sum of enthalpy changes for each reaction