136 CHM General Chemistry II Homework 4, Spring 2009 Dr. Steel
Name
SOLUTIONS
1. A buffer is prepared that has concentration of benzoic acid that is of the concentration of sodium benzoate in the buffer. Find the pH.
pH = pK a + log
[base] where [acid] = 3 [base] 4 [acid] [base] = -log (K a ) + log 3 4 [base] = -log 6.3 10 -5 + log 4 3 = 4.33
(
)
2. A 25.0 mL sample of 0.400 M solution of sodium propanoate has 19.64 mL of 0.200 M HCl added to it. Find the resulting pH. Pro1mol mol molfin [] VT = 44.64 mL 0.0100 -0.00393 0.00607 0.137
+
H3O+
0.00393 -0.00393 0
HPro
0 +0.00393 0.00393 0.0880
+
H2O
Ka, propanoic acid = 1.310-5
pH = pK a + log
[base] [acid]
= -log 1.3 10 -5 + log = 5.08
(
)
(0.137) (0.0880)
3. What mass of BaSO4 will dissolve in 500. mL of water if its Ksp = 1.110-10? [ ]0
[]
BaSO4(s) Ba+2(aq) + SO4-2(aq) --0 0 -x --+x x
K sp = [Ba +2 ][SO 4 ]
-2
+x x
[ ]eq
1.1 10 -10 = (x) 2 x = 1.05 10 -5 M 233.4 g BaSO 4 M 0.500 L = 0.00123 g BaSO 4 1 mol BaSO 4
1.05 10 -5
4. Which system is likely to have more entropy, So? a. 1 L of CO2(g) at 20 C or 1 L of a mixture of CO2(g) and N2(g) at 20 C b. 1 mole of HBr(g) at 5 C c. 1 mole of H2O at 110 C or or 1 mole of HF(g) at 5 C 1 mole of H2O at 40 C
5. A particular reaction has a standard enthalpy of reaction of -84.7 kJ/mol and a standard entropy change of -67.9 J/molK. Determine the range of for temperatures which the reaction will be spontaneous.
o o o Grxn = H rxn - TS rxn = o o J H rxn - 84700 mol T= = = 1247 K o J - 67.9 molK S rxn At all T< 1247, Gorxn is negative, and thus the reaction is spontaneous.
6. Calculate the Horxn, Sorxn, and Gorxn at 298 K for the reaction between butane and oxygen (the equation is UNBALANCED). 2 C4H10(g) + 13 O2(g) These values may help: Substance Hof (kJ/mol) C(g) CO2(g) C4H10(g) H2(g) H2O(g) O2(g) 715.0 -393.5 -126 0 -241.8 0 8 CO2(g) + 10 H2O(g) (balanced)
o H rxn = H fo(products) - H fo(reactants)
So (J/molK) 2.38 213.7 310.0 130.7 188.8 205.1
= [8( -393.5) + 10( -241.8)] - [2( -126) + 13(0)] = -5314 kJ/mol
o o S rxn = S o (products) - S (reactants )
= [8(213.7) + 10(188.8)] - [2(310.0) + 13(205.1)] = 311 .3 J/mol K
o o o Grxn = H rxn - TS rxn
= -5314 kJ/mol - (298...