Solutions Form 3

# Solutions Form 3 - Solutions 1 Solution Kp 1 49 10 8 COCl 2...

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Solutions 1 Solution: K p ± COCl 2 ± g   CO ± g   Cl 2 ± g   1.49 ± 10 8 ± COCl 2 ± g   8.60 ± 10 " 4 atm ± 10 " 4 atm COCl 2 ± g   ± 110 atm Temperature does not come into consideration other than K p depends on temperature. 2 Solution: The second law of thermodynamics says that all spontaneous processes must have S univ ± S sys ² S surr ³ 0 Note: S univ S total G has the opposite sign of S univ so spontaneous processes have G ´ 0. At equilibrium S univ S sys S surr ± 0 and G S univ ± 0 3 Solution: This is a simple weak acid calculation. The initial concentration of HC 2 H 3 O 2 is HC 2 H 3 O 2 ± moles volume ± 15.1 g/60.06 g/mol 0.100 L ± 2.514 M Solve the equilibrium of the weak acid. HC 2 H 3 O 2 ± aq   ¦ H ² ± aq   ² C 2 H 3 O 2 " ± aq   Set up an ICE table: HC 2 H 3 O 2 ± aq   H ² ± aq   C 2 H 3 O 2 " ± aq   I M 00 C " x ² x ² x E " xx x Solve the equilibrium equation

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K a ± H ² C 2 H 3 O 2 " HC 2 H 3 O 2 1.8 ± 10 " 5 ± ¡ x ¢¡ x ¢ 2.514 " x The approximate formula is x ± ± 10 " 5 ±   ± 0.00673 Find the pH pH ± " log ± .00673   ± 2.17 4 Solution: The salt NaF completely dissociates into the ions Na ² and F " . So we must do a weak base calculation for F " . Here is the reaction for the weak base F " . You should be able to figure out this reaction. Bases accept protons so . ..... F " ± aq   ² H 2 O ± e   ¦ HF ± aq   ² OH " ± aq   Set up an ICE table F " HF OH " I 2.16 0 0 C " x ² x ² x E " xx x Solve the equilibrium equation K b ± HF OH " F " K b ± K w K a ± ¡ x ¢¡ x ¢ ¡ " x ¢ The approximate formula is K w K a ± ¡ x ¢¡ x ¢ ¡ ¢ Note that K b is used because this is the equilibrium for a weak base. x ± K w K a ± x ± 5.64 ± 10 " 6 Find the pH
pH ± 14.00 " pOH ± 14 ² log ± x   ± 14 " 5.24 pH ± 8.75 5 Solution: Reduction Cd 2 ² ± aq   ² 2 e " v Cd ± s   Cathode Oxidation Cu ± s   v Cu 2 ² ² 2 e " Anode Look at the half cells. 1. Cu ± s   is dissolving away, Cd ± s   is being formed as a product or plated out. The Cd electrode is gaining Cd metal as the cell operates. 2. Electrons in voltaic cells and electrolytic cells flow through the wire from Anode to cathode. 3. Anions ± "   in solution flow through the salt bridge from Cathode to Anode. Cations ± ²   flow the other way. This means the cations in solution flow from the Anode compartment through the salt bridge then into the Cathode compartment where the Cd 2 ² solution is. The cations don±t flow into the Cu 2 ² solution for this cell. 6 Solution: Remember that with equilibrium expressions the concentrations of pure solids and liquids is 1. So the solid terms don±t contribute to the equilibrium expression. K p ± 1 CO 2 ± g   ± 2.5 CO 2 ± g   ± 0.4 atm 7 Solution: The equation that describes the concentration at various times for a first order reaction is A ± A 0 exp ± " kt   We need to find the rate constant k . We can find k from the half life k ± ln 2 t 1/2 ± ln 2 32.1 min ± 0.0216 min " 1 Find the concentration at time 26.1 min.

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A ± A 0 exp ± " kt   ± 8.92 ± 10 " 2 M ± exp " 0.0216 min " 1 ± 26.1 min ± 0.0508 M Important 1. Ask questions to determine how to solve the problem. Look at the thinking I went
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Solutions Form 3 - Solutions 1 Solution Kp 1 49 10 8 COCl 2...

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