toddychurch

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7. An unknown weak monoprotic acid, HY, is found to be
0.002% ionized in a 1.0 M aqueous solution. What is
the Ka of HY?
A. 2 x 10–3
B. 4 x 10–6
C. 2 x 10–5
D. 4 x 10–10
E. 2 x 10–10

I don't know how to do this. Could someone help me? The explanation provided in the answer booklet is very poor.

Thank you
 

RozhonDDS

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toddychurch said:
7. An unknown weak monoprotic acid, HY, is found to be
0.002% ionized in a 1.0 M aqueous solution. What is
the Ka of HY?
A. 2 x 10–3
B. 4 x 10–6
C. 2 x 10–5
D. 4 x 10–10
E. 2 x 10–10

I don't know how to do this. Could someone help me? The explanation provided in the answer booklet is very poor.

Thank you
Is it B?
 

GeauxTigers

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This could be completely wrong, but I don't think so. I never completely understood these when studying for the DAT, but just memorized how to it.

You have this equation HY --> H(+) + Y(-)

You also have Ka = [H(+)][Y(-)] / [HY] and [H]=[Y]

To find [H] and/or [Y] take .002/100 to get .00002 or 2E-5

Since you're given HY as 1M, now you can solve for Ka.

[2E-5][2E-5] / [1] = 4E-10

Hopefully this helps. Anyone feel free to correct me if wrong.
 
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RozhonDDS

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GeauxTigers said:
This could be completely wrong, but I don't think so. I never completely understood these when studying for the DAT, but just memorized how to it.

You have this equation HY --> H(+) + Y(-)

You also have Ka = [H(+)][Y(-)] / [HY] and [H]=[Y]

To find [H] and/or [Y] take .002/100 to get .00002 or 2E-5

Since you're given HY as 1M, now you can solve for Ka.

[2E-5][2E-5] / [1] = 4E-10

Hopefully this helps. Anyone feel free to correct me if wrong.

That is right. I didn't read the choices real good, but i knew it was 4.0x10 to the something
 

714guy

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toddychurch said:
7. An unknown weak monoprotic acid, HY, is found to be
0.002% ionized in a 1.0 M aqueous solution. What is
the Ka of HY?
A. 2 x 10–3
B. 4 x 10–6
C. 2 x 10–5
D. 4 x 10–10
E. 2 x 10–10

I don't know how to do this. Could someone help me? The explanation provided in the answer booklet is very poor.

Thank you
Percent Ionization is defined as: (molarity of H+ derived from HA/initial molarity of HA) X 100 = %.

.002 % = (molarity of H+ derived from HY/1.0 M (aq) of HY ) X 100

molarity of H+ derived from HY = 0.00002 M

HY --> H+ + Y-

Ksp = [H+][Y-]/[HY] NOTE: [H+] = [Y-] = 0.00002 M

Ksp = [0.00002 M]^2/ [1]

Ksp = 4.0 X 10-10

so the answer would be D. But check whether the question was .002 % or .002 M. if it the later than B is the correct answer.
 
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toddychurch

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You're all correct. In the answer explanation they kept referring to 2 x 10 -5 as [x] and I couldn't figure out where that came from.

Thank you.
 
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