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Finding the PH and POH

  1. Jun 29, 2009 #1
    1. Calculate the PH and the POH of the solution obtained by adding .2g of Solid KOH to 1.5 liters of .002M HCL



    2. Relevant equations



    3.I understand this is a mol/ liter question. but I just don't understand how to transfer .2g of koh to ml and to find the concentration. I know PH= - log [Concentration Someone please help

    I got KOH= 56.1g/mol. so one gram of KOH must = to 56.1 moles. If i multiply by .2 I get that .2g of KOH =11.22g
     
    Last edited: Jun 29, 2009
  2. jcsd
  3. Jun 29, 2009 #2

    symbolipoint

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    Your mass to moles for KOH is correct. Next, find the excess number of moles of either KOH or HCl, and this will tell you if the solution is acidic (for which you find pH), or alkaline (for which you find pOH). To find the corresponding other pX value, you use pH+pOH=14
     
  4. Jun 30, 2009 #3

    Borek

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    No, that's not OK. Why do you multiply mass of the KOH by the molar mass of the KOH? 11.22 is not number of moles, besides, it is not in g - it is in g2mol-1.

    Check this calculation of pH of mixture example.
     
  5. Jun 30, 2009 #4

    symbolipoint

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    My mistake is the first sentence:
    Correct would be (0.2 grams KOH)/(56.1 grams KOH / mole KOH)= theMolesOf KOH.
    Read that text carefully; that is supposed to be a division of 0.2 by 56.1
    Note the slash mark (actually two of them. The first is for the numbers, the second is for a ratio)
     
  6. Jul 5, 2009 #5
    So 0.2g KOH/56.1g/mol KOH = .003 mol and 0.002M HCL * 1.5L HCL = 0.003 mol.

    So 0.003 mol KOH - 0.003 mol HCl =0, so the pH has to be 7, right?

    But what if there was still some of a reagent left? Then you would need to calculate the volume of KOH, correct? And how would you do that when you don't have the density (D=mV)?
     
  7. Jul 5, 2009 #6

    Borek

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    Don't round down intermediate results.

    You don't need density here - you know there is 1.5L of solution. Volume almost doesn't change when adding so small amount of solid reactant.
     
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