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Posted by on Sunday, May 15, 2011 at 8:02pm.

Calculate the pH at the equivalence point for the titration of a solution containing 1250.0 mg of hilariamine (MW = 92.5 g/mol, Kb = 8.1×10−4) with 0.1000 M HCl solution. The volume of the solution at the equivalence point is 175.0 mL.

(I got the answer to be 12.0208 but it seems wrong because pH isn't suppose to be that high)

  • chemistry - , Sunday, May 15, 2011 at 8:21pm

    What about 6.01?
    This is an amine so I'm calling it RNH2. At the equivalence point with HCl the salt will be RNH3Cl and the pH will be determined by the hydrolysis of the salt.
    RNH3^+ + H2O ==> RNH2 + H3O^+

    Ka = (Kw/Kb) = (RNH2)(H3O^+)/(RNH2).
    Kw/Kb = 1E-14/8.1E-4 = 1.23E-11
    1.23E-11 = (x)(x)/(salt)
    (salt) = M = moles/L and moles = g/molar mass
    moles = 1.25/92.5 and volume = 0.175; therefore, 1.25/92.5/0.175 = about 0.077
    1.23E-11 = (x)(x)/(0.077)
    Solve for (H3O^+) and convert to pH. You need to work through it without estimating here and there as I did.

  • chemistry - , Sunday, May 15, 2011 at 9:03pm

    i worked through it just don't know it i did it right...

  • chemistry - , Sunday, May 15, 2011 at 9:36pm

    And your answer should be 6.01

  • chemistry - , Monday, May 16, 2011 at 5:14pm

    i checked through my work and I found out i didn't do square root for the x^2....thats why my answer was double 6.011.

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