I worked the first part and got 0.01 M for (H2X) which is close enough to 0.00995 to be the same thing. I think then you should have done this.
(H2X) = 0.01 M = 0.01 mols/L
0.01 molar x 0.250 L - 0.0025 mols.
then mols = g/molar mas
0.00250 = 0.225/molar mass and I get a molar mass of about 90 or so which sounds better than the 5. Check me out on that.
For part 2, what do you mean by BCA? I'm interested in what that stands for.
I tried to estimate k2 by
(H^+)^2 = k1k2 but that gave a k2 of ?? x 10-15 which I think is out of the ball park. I expect you will need to use the more complicated formula to determine k2.
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
5 answers
ohh god I'm silly i divided the L instead of multiplying it=\ thanks a lot BCA I'm not really sure what it stands for but its when you add a strong acid to a weak base or a strong base to a weak acid its kinda similar to a ice table but your using mol instead of molarity and you're subtracting by the lowest mol instead of the complicated stuff that you have to do to ice table i'm thinking it deals with buffer since I'm learning buffer atm here a sample equation that requires BCA ( H+ + H2NNH2---> H2NNH3)
first to solve part b i think you guy shold use the equation pH= -logKa + log [A]/[HA]. remember that the molarity at the second equilibrium point will be just the same as the first one or their ratio still the same. Now you just got all the information plug in and solve for ka at second point.
BCA stands for Before Change After
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2 answers