<p>for the temp question is the answer 0 K? also for the reactions, is one of the answers acid rain....and another Ag is left in the filter paper?</p>
<p>HOW IS THE CONCENTRATION for the weak acid .1 I don't understand... ***</p>
<p>Oh well, I could have sworn that dH and dS were independent of temperature, but that might be working opposite against what I was trying to prove. Have fun discussing this test.</p>
<p>For 6f and g, did you guys say</p>
<p>f. decrease because Lechatlier says increasing temp in an exo reaction shifts it to the right, making more product, and since product is in the denominator, ratio is smaller</p>
<p>g. catalysts dont change ratio, only make the reaction reach equiblirium faster</p>
<p>^^ I said the ratio increased because increasing the temperature here shifts it to the left.</p>
<p>Ptang, multiply the molarities and milliters together to get millimoles, and you get 10 mmol of HF and 6 of OH, so OH is used up, set up an ice chart, subtract 6 from HF and add 6 to F-, and then divide 6 mmol by (15+25 mL) to get .15 M of the weak acid</p>
<p>I thought adding heat in an exothermic reaction shifted it left. But I was never good with that stuff. And for 5, I think its cler to purple.</p>
<p>"Oh well, I could have sworn that dH and dS were independent of temperature."
No, you're right.</p>
<p>Increasing temperature increases k(endothermic) more than k(exothermic), so the equilibrium shifts in the endothermic direction, which formed more reactants in this case, I think?</p>
<p>Hmm. I don't think I did very well on the weak acid question. I can't remember if I got an acidic or basic pH...</p>
<p>I just looked in princeton review...
pH = pKa + log{[A-]/[HA]}</p>
<p>where [HA] = molar concentration of UNDISSOCIATED weak acid (M)
[A-] = molar concentration of conjugate base.....</p>
<p>Therefore you cannot take the answer from part c which is the number of moles of HF(aq) remaining in the solution and divide it by the new volume which is 40 mL</p>
<p>Am I right?</p>
<p>Crap. I was thinking exo as heat as a product, wasnt quite sure which way. Why didnt I just think of endo with heat as a reactant, that shifts right so much more obviously....guh.</p>
<p>"Oh well, I could have sworn that dH and dS were independent of temperature."
They are independent, but dG is not</p>
<p>I don't remember what c, d, and e asked.... can someone remind me?</p>
<p>Unfortunately, Ptang, that's not correct. Mike's explanation is correct.</p>
<p>Also, reaction shifts left. Increased heat favors the endothermic process. In exo process, heat is product. Raising temp is like increasing concentration of product, therefore shifts left to use it up.</p>
<p>A volume of 15 mL of 0.40 M NaOH is added to 25 mL of 0.40 M HF</p>
<p>d. Find moles of HF remaining in solution
e. Find molar concentration of F-
f. Find pH of solution</p>
<p>wow, i thought i screwed up a lot, but only a couple of points off so far!</p>
<p>OK someone please explain y Mike is right... I understand what he did, but if you take the moles of HF remaining shouldn't you also take the moles of OH remaining, which would be 0 therefore making it wrong????
anyone wanna tell me why???</p>
<p>You use up the moles of OH, there is none. That has no effect on the moles of F- formed. How would OH being 0 make it wrong?</p>
<p>yeah I see now... crap I lose another point</p>
<p>I am very upset because parts G And F for the very last question was not printed on my sheet. Either that or I somehow did not see them which I find very unlikely. Did anyone else not have a part G and F for question 6?</p>
<p>Also for the 2nd to last question, instead of</p>
<p>i.) 5MV ii) 5(55.85)MV iii)5(55.85)MV/g</p>
<p>I did</p>
<p>i.) MV/5 ii) (55.85)MV/5 iii)(55.85)MV/5/g</p>
<p>Will I lose all the points or get partial credit?</p>
<p>I didnt see it I don't think it was printed on my sheet either I don't really think I could have missed them</p>