anonymous
  • anonymous
Assume we start with 0.025M NaOH. 100mL of this 0.025M NaOH is poured into 0.5g Ca(OH)2, and this will eventually form a saturated solution. Then, a 25mL aliquot of this solution is taken. What is the concentration of NaOH now? Attempt: 0.025mol/L * (0.1L) = 0.0025mol of NaOH in total Then when you take 25mL from this 100mL part of it, we get: 0.0025mol / 0.025L = 0.1M NaOH So, when you take 25mL portion of this solution, we have concentration of 0.1M NaOH now...?
Chemistry
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katieb
  • katieb
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anonymous
  • anonymous
Are you sure the question asks what the concentration of NaOH is and not OH minus? Becuase NaOH dissociates completely so it's concentration should be close to zero and it looks like since you're adding OH minus from the calcium hydroxide that they wan't you to find the OH final after addition of CaOH2
anonymous
  • anonymous
Sorry, I meant the concentration of OH- coming from NaOH
anonymous
  • anonymous
That's what I thought. So you start with 100ml of a .025M solution of NaOH. You calculated mols but you forgot to add the moles from CaOH2 before taking the molariity. Convert the 5 grams of CaOH2 into moles. We know that there are twice as many OH's as CaOH2's so multiply whatever number of moles you get by two. Add the amount of moles you just found to the moles you found from the NaOH, then divide by the .025L.\[{{5gCaOH _{2} \over 57molarmass} \times {2moles OH ^{-} \over 1moleCaOH _{2}} + {.025MNaOH \over .1L} \over .025L} =[OH ^{-}]\]

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anonymous
  • anonymous
So I still do that calculation of this: 0.025mol/L * (0.1L) = 0.0025mol of NaOH in total?
anonymous
  • anonymous
So what I do is get the moles of Ca(OH)2, and get the moles of NaOH by multiplying the concentration by 0.1L. Then, I add up mol OH- from both sources, and divide by 0.025L (which is 25mL)?
anonymous
  • anonymous
Perfect.
anonymous
  • anonymous
Actually I think this is incorrect. I apologize. In my previous formula you should divide by .1L to get the total concentration in the beaker. Dividing by .025L is incorrect because then it's like saying we took the amount of moles we had and stuck it in 25ml which is not the case. So in my previous equation divide by .1L. That will be your OH concentration. It doesn't not matter if you take a 5ml aliquot or 250ml aliquot out, the concentration will not change.

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