CHEM132 Exam3 review PDF

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Review material for Test 3 for Dr. Kokhan...

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3/31/2016

When 0.0400 moles NaF reacts with 0.0200 moles HCl in a total volume of 40.0 mL, the resulting solution is: I. A buffer II. at the equivalence point III. neutral NaF + HCl → Na+ + Cl- + HF Initial 0.04 0.02 0 Equil. 0.02 0 0.02

Have both HF and F- => buffer and not at equivalence point pH = pKa + log ([F-] / [HF]) = pKa + log 1 = pKa pKa of HF is something acidic so III is wrong

a) Buffer should contain a mixture of a weak acid (HA, red and blue) and its conjugate base (A-, red only). Only (A) and (D) are buffers as (B) is weak base and (C) is weak acid b) for buffers you have to use H-H. equations: (A) pH = pKa + log ([base]/[acid]) pH = 4.35 + log (3/3) = 4.35 (D) pH = 4.35 + log (5/3) = 4.57 (B) you approach this problem like any other “pH of weak base” problem: 𝑥∗𝑥 𝐾 𝐻𝐴 [𝑂𝐻 − ] = 𝐾𝑏 = 𝑤 = 2.22𝑥10−10 = ≈ [𝐴− ] 0.1−𝑥

𝐾𝑎

𝑥2 0.1

[OH-] = x = 4.71x10-6M => pOH = -log(4.71x10-6) = 5.33 pH = 14 – pOH = 8.67 (C) you approach this problem like any other “pH of weak acid” problem: 𝐾𝑎 = 4.5𝑥10−5 =

c) C (weak acid) -> A (buffer) -> -> D (buffer with more base) -> B (base)

𝑥∗𝑥 𝐴− [𝐻3𝑂 +] = [𝐻𝐴] 0.1−𝑥

≈

𝑥2 0.1

[H3O+] = x = 2.12x10-3 M => pH = -log(2.12x10-3) = 2.67

d) B – at the equivalence point of weak acid titration all weak acid is converted to conjugate weak base

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3/31/2016

What prediction can we make about the pH of a solution formed by adding 25.0 mL of 0.15 M NaOH to 25.0 mL of 0.15 M HF? I. II. III. IV.

Acidic Basic Neutral pH = pKa NaOH + HF → Na+ + H2O + F-

We have the same number of moles for NaOH and HF, therefore, both will be consumed in the reaction and we’ll have only F-. F- is a conjugate base of a weak acid, therefore, it must be a weak base

Weak base in water will give basic pH (II) From Henderson-Hasselbalch eqn: pH = pKa + log (base / acid) pH = pKa only when [base] = [acid]. We don’t have any HF left, therefore, IV is wrong

What is the correct relationship between Ksp and molar solubility s for Al(OH)3? s s 3s Al(OH)3 → Al3+ + 3 OHKsp = [Al3+][OH-]3 Ksp = s * (3s)3 = s * 27s3 = 27s4

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3/31/2016

The Ksp for AgBr is 5.0 x 10-13. What is the molar solubility of AgBr in water?

AgBr → Ag+ + BrKsp = [Ag+][Br-] = s*s = s2 = 5.0x10-13 s = 7.1x10-7 M

What is the molar solubility of silver(I) chloride (Ksp = 1.8 x 10-10) in 100.0 mL of a solution that already contains 2.50 M sodium chloride? AgCl → Ag+ + ClKsp = [Ag][Cl-] [Ag+] = s [Cl-]

(coming only from AgCl)

= s + 2.5 (s from AgCl and 2.5M from NaCl)

Ksp = s * (s + 2.5) ≈ s * 2.5 s = Ksp / 2.5 = 1.8x10-10 / 2.5 = 7.2 x 10-11 M

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3/31/2016

For the following reaction, we can see that entropy is ___, which means that if this process occurs, it must be____-othermic. 2SO2(g) + O2(g) ↔ 2 SO3(g) 3 mols of gas on left, 2 mols on right => decrease in entropy Reaction with oxygen is combustion => heat release => exothermic

Consider the reaction A + B AB At 700 K, ΔG° = 3.4 kJ and ΔG = -0.7 kJ. What is favored at standard state concentrations at equilibrium at 700 K, and which direction does the reaction proceed to reach equilibrium? Products are favored if ΔG < 0 Reactants ΔG > 0

True for any set of concentrations (i.e. Q)

ΔG = ΔG° + RT lnQ At standard state Q = 1 (since all concentrations = 1) ΔG = ΔG° Sign of ΔG° tells us what is favored at standard state concentrations So, for this reaction: reactants are favored at standard state concentrations ( ΔG° > 0) products are favored at conditions where ΔG = -0.7 kJ

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3/31/2016

Determine the free energy of reaction, ΔG, for the decomposition of SO3 to form SO2 at 500 K when the initial pressures are PSO2 = 0.500 atm, PO2 = 0.010 atm and PSO3 = 0.10 atm. 2 SO3(g) 2 SO2(g) + O2(g) ΔG° = +104.5 kJ at 500 K ΔG = ΔG° + RT ln Q Q = PSO22 * PO2 / PSO32 = 0.52 * 0.01 / (0.1)2 = 0.25 ΔG = 104.5 kJ * 103 J / kJ + 8.31 J/K * 500K * ln (0.25) = 98700 J = 98.7 kJ

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