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PH question

  1. Oct 12, 2004 #1
    Hi ppl, I got a problem with this question

    Calculate the pH of 2.00dm3 of an aqueous solution which contains 12.4g of CH3NH2 and 13.5g of methylammonium chloride, CH3NH3+Cl-. The Pka of methylammonium chloride is given as 10.8

    Thanks a lot for any help,
  2. jcsd
  3. Oct 13, 2004 #2
    pH of a solution

    we will apply Henderson-Hasselbach equation here
    which says that pH = pKa + log [salt] / [acid]
    salt is CH3NH3Cl and acid is CH3NH2
    to determine concentartion we have to calculate the Molarity of each.
    mol wt of salt = 67.521, mol wt of acid = 31.068
    therefore moles of salt = wt / mol wt = 13.5/67.521 = 0.199
    moles of acid = 12.4 / 31.068 = 0.399
    as the solution is 2dm3 that means 2 litres, hence molarity of each will be
    salt = 0.199 / 2 = 0.099M, acid = 0.399 / 2 = 0.199M
    putting the values in equation
    pH = 10.8 + log 0.099 / 0.199
    pH = 10.5
  4. Oct 13, 2004 #3


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    Gold Member

    Your approach seems to be okay, but you can omit the volume as they are present in the same 2 liters of solution, so just divide the mole values of base and acid each.
  5. Oct 14, 2004 #4

    any how it won't effect the answer... it was just to explain the basis of henderson hasselbach equation and what exactly means by concentration so it can be used in varying occasions
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