Chemistry Homework Help

[NoSwtTea4U]

Guys, I figured there's some people on here that have/have had CHEM 1212 in college and I'm having a little trouble lately. Not just me, but 22 out of 41 people are failing the 2 classes my teacher teaches. My class size is 16-started at 33 :frustrate , but nevertheless...If anyone would care to help me on exactly how to go about doing these, that would be great, thanks!

 

[Enigma76]

2 is easy, just use (molarity acid)(volume acid) = (molarity base)(volume base)

3 is a bti more complicated. convert KOH to moles, and you know that since KOH is a strong base and HClO4 is a strong acid, every mole of acid added will subtract an equal amoutn of moles from the base (they will neutralize each other). you have to work in moles for this question.

 

[NoSwtTea4U]

2 is easy, just use (molarity acid)(volume acid) = (molarity base)(volume base)

3 is a bti more complicated. convert KOH to moles, and you know that since KOH is a strong base and HClO4 is a strong acid, every mole of acid added will subtract an equal amoutn of moles from the base (they will neutralize each other). you have to work in moles for this question.

Thanks for #2-I was making it more complicated than it was. As for #3, I'm still not getting it I realize for "a)" You have .0045M for the acid and .006M for the base...Where to from here?

 

[jmh80]

These are all strong acid/strong base chem problems. Easy stuff man. Just wait till you get to buffers.

I took all the way through organic 1&2, analytical, and physical chem.

Enigma is right. For #2, multiply the molarity of the base times the volume given divided by 1000 (to convert to liters). This gives total moles of base. Then simply divide by the molarity of the acid (mols/mols/L) to get the volume required. Pretty simple.

3 is more difficult. You have the number of mols of base, 30/1000 * 0.2. I would then calc. the mols of acid the same way for each. Subtract the mols base from acid. When it is positive, the solution is basic (pH above 7), when negitive it is acidic, and when equal it is neutral (pH = 7). Once you get the mols excess, I think you can just divide that by the total volume to get into concentration of H+ or OH-. The log base 10 would then be the pH if it is acidic. (I'm pretty sure pH = log base 10 of the concentration of H+ in mol/L - someone check me on that.) If it is basic, you can calc. the pOH the same way. pH is then 14 - pOH.

I think you picked a very good board to ask a question like this, we've got some damn good chem people on here!