<p>When an ideal gas is allowed to expand isothermally, what does it mean in terms of E, q, and w?</p>
<p>bump... maybe i need to elaborate on question: well, you know the equation for internal energy, E=q+w. when the gas expands isothermally, how do E, q, and w change? (as in, does one of them become 0)</p>
<p>also another stupid question i wasnt sure about: calculate the pH of a (2)(10^-11)M HCl solution.</p>
<p>thanks</p>
<p>for a 2 x 10^-11 M HCl solution, HCl dissociates extensively so for each HCl u get one H+ therefore u have 2 x 10^-11 H+ ions.... to find pH, take the -log(2x10^-11) and thats ur pH!! voila!! i would do it cept i dont have a calculator...</p>
<p>um yeah. thats what i thought at first, which would give you a pH bwteen 10 and 11. However, conceptually, there's no "extra" OH- in the solution except those from water; in fact, there are actually extra 2*10^-11 M H+ ions (besides those from water) - thus, it makes no sense that the pH would be basic. so what is the pH?</p>
<p>Yeah...that gives you a pH of 10.7, which makes no sense because HCl is a strong acid. Perhaps the concentration is incorrect?</p>
<p>Do an equilibrium reaction.
HCL <=> H+ + Cl-
I 2<em>10^-11 M, 0, 0
C -x, +x,+x
E 2</em>10^-11 - x, x, x</p>
<p>Ka = x^2/(2<em>10^-11-x) = x^2/(2</em>10^-11) because x is negligible</p>
<p>not sure what the Ka for HCl is, but solve for x and take the -log and you get it (I think)</p>
<p>Yeah...that gives you a pH of 10.7, which makes no sense because HCl is a strong acid. Perhaps the concentration is incorrect?</p>
<p>when u think about it..it makes perfect sense..</p>
<p>in normal water..the [H+] is 1.0 x 10^-7....so when u have a soln. with a [H+] of 1.0 x 10^-11, it has less H+ than water..thus having more OH- than the autoionization of water would dictate...its a basic solution
so in fact it is not really HCl but a weak base with Cl- ions floating around!!!</p>
<p>Some website I found gave 1x10^6 as the Ka of HCl, which would make my answer pH = 2.349485, which seems really low considering the little amt of the HCl used...but I think I was on the right track.</p>
<p>well, i saw this problem at a chem competition. The correct answer was 7. i put 10.7 after doing the -log thingy
i was extremely puzzled by why the math doesnt give the right answer. i thought that since HCl is a strong acid, it dissociates almost completely and you can just do -log[H+]</p>
<p>ok with the 2x10-11 M HCL solution.
just went over this a week ago.</p>
<p>Equilibrium reaction for water
H2O <--> H+ + OH-
in normal H2O, there are 1.0E-7 M H+, and 1.0E-7M OH-
also, (H+)x(OH-)=1E-14 (Ka of water)
if H+ = 1E-11, OH- = 1E-3
-log(1E-3)=x=POh
14 = Ph + POh
14-x = Ph</p>
<p>the solution is basic. as well it should be.</p>
<p>HCL doesn't have a Ka, as arti said. it is a strong acid, and therefore fully dissassociates. There is no equilibrium whatsoever w/ HCl.</p>
<p>For the thermo question,</p>
<p>I think delta E and delta q will be both 0, because of no temperature change. So the only change in volume will be due to work, so w will have a value on it.</p>