DAT destroyer chem questions

[20090900]

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I have the 2012 version but I don't think the questions are different/the numbering might differ +/- 1 or 2 spots. I am having trouble understanding these questions even with the solution and any help would be most appreciated.

#75: Find the pH of a 0.040M NaCN solution. Given Ka of HCN = 5.0 x 10^-10
---> how do you know to calculate the Kb using the Ka? This is what I don't understand about the question. When I did it; I just used Ka right away, solved for x, and found pH..but this is obviously wrong. (Ans: pH = 11)

#89: A gas is compressed from 40 to 16 liters at a constant pressure of 4.0 atm. If 10kJ of energy is released as heat, find the work done for this process.
---> so what I don't understand is why work is (+) and not (-)...the equation (Work = -PdeltaV, and E = Q + W or whatever) confuses me and I'd rather just try and use logic. I know that if the SYSTEM does work on the SURROUNDINGS, W is NEGATIVE b/c the system loses energy. Work is POSITIVE when the SURROUNDINGS does work on the SYSTEm b/c the system gains energy. So in this question, if you're compressing a gas that obviously requires work from the system to be done on the surroundings, and even to back this up, 10kJ of energy is lost. So why is work positive and not negative?....I thought this was a fairly simple concept but my thinking is obviously wrong.

Thank-you! Also I will most likely have other questions so I really appreciate the help + patience.

 

[Dgeorg6]

for #75 Kw = Ka x Kb and Kw is always 1x10^-14 at a 25 *C

so to find Kb you do Kw / Ka

and you know to find the Kb because it is asking the pH of NaCN (which is a base) not HCN (which is an acid), so therefore you have to find the pOH 1st then pH

#89

The gas is compressed which means you lose Pressure but you dont even need to know that. Your Delta Volume was -24 because the change was 40 to 16 so there forever -P X -Delta Volume = + work

 

[joeeeey3]

75. Kb can be found using Kw = (Ka)(Kb) and Kw is 1x10^-14 at standard conditions, so use Kb to find the [OH-] concentrations -> pOH then subtract from 14 to get pH

89. 10kJ is heat not work, and they put it there to just to confuse you. volume decreased from 40L to 16L (surrounding did the work cuz surrounding pushed down the system) so the work is +ve. then just use formula W = PV = 4 atm (24L) = 96 joules