How much OH-?

[DingDongD]

If there is 2.00*10^-7 amount of H+, how much OH- is there?

So, I did it by this way:

H20-> H+ + OH-
I 2.0*10^-7 1.0*10^-7
C -x -x
E 2.0*10^-7-x 1.0*10^-7 -x

(2.0*10^-7-x) (1.0*10^-7 -x) = 1.0*10^-14

For x, I got 3.89*10^-8, so the total amount of OH- is 6.11*10^-8.

However, when I just do:

1.0*10^-14 = (2.0*10^-7) (X)

X = 5.0*10^-8 amount of OH-

So, which method is correct? I am guessing the first one.

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[erythrocyte666]

the second one is correct.
in the first one, you can't assume OH- has initial concentration of 1 x 10^-7. the auto ionization property of water dictates that the product of H+ and OH- is always 10^-14, so if you're given the conc for H+, then you just need to do your second method. if you assume initial OH- is 10^-7, the product of H+ and OH- becomes 2 x 10^-14, which ain't happening at room temperature

 

[DingDongD]

But, its displaced out of equilibrium originally in the first method. So, q>k in this case. In order to return to k, shouldn't there be a change, which was x?

 

[DingDongD]

Nm. got it!

 

[SN2ed]

Thread closed. These types of questions belong in the MCAT Study Q&A.