C_Sagan_Returns Posted November 23, 2008 Share Posted November 23, 2008 Check this out. I've had the solution for weeks and I still can't figure out how to get these answers without pluging in all seven equations and seven unknowns (or whatever it is) into a calculator. I tried every possible combination and my professor doesn't help very much, he'll never show you step-by-step how to do something...he just repeats the theory to you over and over. I'm guessing it's just my poor algebra skills getting in the way. Can someone please show me how to go through part (2) of the solution (it's not shown step by step!!). Thanks, CSR Link to comment Share on other sites More sharing options...
C_Sagan_Returns Posted November 25, 2008 Author Share Posted November 25, 2008 For those of you trying to help me out, I finally figured out how to set everything up (thanks to my prof). Starting with: Moles Acf = moles Aci + moles OH Moles HAcf = moles HAci – moles OH Substitute: “(0.1M)Vol OH” for “moles of OH” in both equations above. Then, substitute: “(Aci)Vol Acid” and “(HAci)Vol Acid” for “moles of Aci/HAci” respectively. This will leave you with: Moles Acf = (Aci)Vol Acid + (0.1M)Vol OH Moles HAcf = (HAci)Vol Acid - (0.1M)Vol OH These two equations can be plugged into the Henderson-Hasselbach Equation yielding: pH = pK + log[ ( (Aci)Vol Acid + (0.1M)Vol OH ) / ( (HAci)Vol Acid - (0.1M)Vol OH ) ] Once you plug in all the numbers and do some math you get: (.4295M)Vol Acid = (.5365M)Vol OH Then, substitute: “(0.05L – Vol Acid)” for Vol OH This leaves you with one unknown…Vol Acid = 0.028L or 28ml That took entirely too long! I guess it was just my algebra. Thanks anyways guys, CSR Link to comment Share on other sites More sharing options...
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