physical chem.docx - Exercise 9.2 Question 1 PH=-log(0.0058...

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Exercise 9.2 Question 1 PH= -log (0.0058) = 2.2 Question 2 [Ba][F - ] 2 γ Ba [Ba] γ F [F - ] 2 γ Ba [s] γ F [2s] 2 0.44(6.8x10 -3 ) X 0.83(6.8X10 -3 ) 2 X 4 4.6 X 10 -7 Exercise 9.3 Question 2 Solubility equilibrium AgI (s) ↔ Ag + + I - Ksp = [Ag + ] [I - ] If the solubility of AgI is x M, [Ag + ] = [I - ] = xM Ksp = x 2 x = Ksp = 8 × 10 17 = 8.9 × 10 9 So the solubility of the AgI in pure water is 8.9 × 10 9 Considering activities Ksp= [Ag + ] γ Ag + [I - ] γ I - If the solubility of AgI is xM, [Ag + ] = [ I - ] = xM Ksp = x 2 γ Ag + γ I - x= γ Ag + ¿ γ I ¿ Ksp ¿ ¿
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γ represent the activity coefficient γ = 10 [ 0.51 z 2 4 1 + a 4 305 ] µ= + ¿ Na ¿ ¿ ¿ NO 3 ¿ (− 1 ) 2 ¿ 1 2 ¿ from 0.008M NaNO 3 , [Na + ] = 0.008M and [NO 3 - ] = 0.008M µ= 1 2 ( ( 0.008 ) ( 1 ) +( 0.008 )( 1 ) ) = 0.008M γ Ag + = 10 ( 0.51 (+ 1 ) 2 0.008 1 + 250 0.008 305 ) = 0.907 γ I - = 10 ( 0.51 (− 1 ) 2 0.008 1 + 300 0.008 305 ) = 0.908 x ¿ 8 × 10 17 ( 0.907 )( 0.908 ) = 9.9 × 10 9 Exercise 10.1 Q no. 4 C3H3O3 - C3H5O3 - C6H12O6 C6H17O7 - Step 1 balance oxygen by adding H2O
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C3H3O3 - C3H5O3 - H2O + C6H12O6 C6H17O7 - Step 2 balance hydrogen by adding H+ 2H + + C3H3O3 - - C3H5O3 - H2O + C6H12O6 C6H17O7 - + 3H + Step 3 balance charge e by increasing electron 2e - + 2H + + C3H3O3 - C3H5O3 - H2O + C6H12O6 C6H17O7 - + 3H + + 2e Overall reaction C3H3O3 - + C6H12O6 + H2O C3H5O3 - + C6H17O7 - + H + Exercise 10.2 Q no. 1 V 2+ + 2e - V E ox = x H 2 2H + + 2e - E red = 0V E = E 0 – 0.0592/n log (10
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