Question

1. consider an aqueous solution that is prepared with pure water and an excess baf₂(s).

chem 203 - tutorial 5: salts & thermochemistry
baf₂(s) ⇌ ba²⁺(aq) + 2f⁻(aq)
data:
kₛₚ baf₂ = 1×10⁻⁶
kₐ hf = 6.3×10⁻⁴
would the following changes increase, decrease, or have no effect on the amount of baf₂ that dissolves in the solution? write the proper word or expression in each box. 2 points
increase / decrease / no effect
(the amount that dissolves)
a. add some kf to the container.
b. add some kcl to the container.
c. add some water to the container.
d. add some hcl to the container.
written - answer questions (hand - graded) - write your final answers in the answer box, including units for numerical answers. show all your intermediate steps of calculations in the work area for full points. work outside the answer areas will not be graded.

1. one of the reactions used to inflate automobile airbags is the decomposition of guanidine nitrate (below) to produce nitrogen gas.

c(nh₂)₃no₃(s) → 3h₂o(l) + 2n₂(g) + c(s) δhᵣₓₙ = - 470.0 kj/mol c(nh₂)₃no₃
a. how much pv work is done when 0.125 mol of guanidine nitrate decomposes completely at 20.0°c and an ambient pressure of 1.20 atm? report your answer in units of j 1.5 points
answer:
b. how much heat is exchanged when 0.125 mol of guanidine nitrate decomposes? report your answer in units of j or kj 0.5 points
answer:

Explanation:

Question 3

a. KF provides F⁻ ions, shifting equilibrium left (common ion effect).
b. KCl has no common ions; ionic strength change is negligible for solubility here.
c. Adding water dilutes the solution, shifting equilibrium right to dissolve more BaF₂.
d. HCl provides H⁺, which reacts with F⁻ to form HF (weak acid), reducing [F⁻] and shifting equilibrium right.

Step1: Find Δn_gas

Δn_gas = moles of gaseous products - reactants = (2) - 0 = 2 mol per mol of reactant.

Step2: Calculate Δn for 0.125 mol reactant

Δn = 0.125 mol × 2 = 0.25 mol.

Step3: Convert T to Kelvin

T = 20.0 + 273.15 = 293.15 K.

Step4: Calculate PV work

Work \( w = -P\Delta V = -ΔnRT = -(0.25)(8.314)(293.15) \)

Step1: Relate heat to ΔH

Heat exchanged (q_p) = n × ΔH_comb.

Step2: Calculate q

q = 0.125 mol × (-470.0 kJ/mol) = -58.75 kJ = -58750 J.

Answer:

a. decrease
b. no effect
c. increase
d. increase

Question 4a