a) If it was a strong acid, the pH would have been 7 at the equivalence point (40 mL)

b) HA + OH- --> H2O + A-

c) .25 x .04 (1mol HA/1mol NaOH) = .01 mol HA

d) mol = g/mw
.01 mol = 1.22/ mw mw = 122 g/mol

e) Ka = (x^2)/.200 x = .0035 pH = -log .0035 = 2.45

f) pH=4.2 + log ((0.0075)/(0.01-0.0075)) H+ = 2.1 x 10^-5 M

2)
a) 6.09 x 10^-3 (PV= nRT)

b) 1.4-.2 = 1.20 atm

c) .2 atm (5/1) = 1.0 atm (when working on this problem I created the equation found in part d during the process, imagine my surprise when I saw what the question for part d was. This probably means they were expecting you to do it without the full equation.)

d) C3H8 + 5O2 --> 3CO2 + 4H2O
We know this because we know since O2 was left over that the hydrocarbon was the limiting reagent, therefore since .2 atm of hydrocarbon was consumed and .6 atm of CO2 was formed, then it must be a 3:1 CO2 to hydrocarbon ratio meaning the hydrocarbon must have 3 C. Repeat for the 8 H

e) .00609 * (44g/mol) = .268 g (As long as you used the hydrocarbons mw you got for part d you should get full credit)

f) the CO2 reacts with H2O to form H+ and HCO3-, pH < 7

3)
a) products - reactants = 127.2
b) bonds broken (reactants) - bonds formed (products) = 49 (note that some students do this in reverse and get -49, the answer is definitely +49)

c) temp goes down because + delta H means endothermic (if you got -49 for part b you should get the credit if you put temp goes up because - delta H means exothermic)

d) ln[a]t - ln[a]0 = -kt k= .094 min^-1

e) Rate = (.094)(2.35x10^-3)
Rate = 2.21 x 10^-4 M/min

f) curve, ln[A] vs time is a straight line for first order, 1/[A] vs time would be curved

4)
a) SrCO3 + 2H+ --> H2O + CO2 + Sr2+ , gas given off, bubbles form
b) 2Mg + O2 --> 2MgO, 0 --> +2
c) Ni2+ + 2OH- -> Ni(OH)2 , Cl-, 2 Cl- for ever NiCl2 unit

5) (Note: Always hard to tell what college board is going to be looking for on these.)
a) Both are nonpolar and therefore the primary IMFs are dispersion forces (LDFs) in both of them. Since I2 is more polarizable due to its larger electron cloud, it has stronger LDFs, making it require more energy to overcome the