February 22, 2017

Homework Help: chemistry

Posted by timmathy on Sunday, March 2, 2008 at 2:09pm.

225 mg sample of diprotic acid is dissolved in enough water to make 250mL solution.the pH of this solution is 2.06. a saturated solution of calcium hydroxide Ksp=1.6E-6 Then the saturated solution of calcium hydroxide is prepared by adding excess calcium hydroxide to pure water and then removing the undissolved solid by filtration. Enough calcium hydroxide solution is added to the solution of the acid to reach the equivalence point.

The pH of the second equivalence point is 7.96. Ka1 is 5.90E(-2). Assume that the volumes of the solutions are additive, all solutions are at same condition, and Ka1 is at least 1000 times greater than Ka2
what is the molar mass of the acid ?
and what is the second dissociation constant for the acid?

i calculkated the molar mass but it came out to be reallie wierd can some check my math to see where i'm off

225 mg of H2X(X being the unknown) ph=2.06 [H]= .00871 i set up a ice table
H2X---------------------->H+ + HX
z(the unknown molarity) 0 0
-x +x +x
x=[H] so x is .00871 i plug in the variable then [.oo871][.oo871]/[z-.oo871]= 5.9E-2(the ka1)i cross multiple and got .009995 so i divided it by .25 which is the L of the solution to get .039981mole
i divided .225g into the mole i got but i got 5.62761g/mol which is kinda wierd shoulkd it be large=\

and on solving the ka2 wouldnt i have to do a BCA table then ice table using the moles i got from the ka1 ice table? and would pure water have any effect on it?

i calculated the

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions