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#### galpinj

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##### Question about pH with Water
« on: September 29, 2016, 10:42:46 PM »
Hello,

I worked through a titration problem recently, and the weak base (F-) that resulted had a concentration of: [OH-] = 8.6 x 10-7

My question is, given that this number is very close to the concentration of hydroxide ion in pure water (1 x 10-7), would I have to add these two concentrations together to determine the [OH-]?

In the example, the concentration of OH- in water is ignored, but I feel like it should be included.

Finally, how would this effect (OH-)(H+)=1 x 10-14? Would I add the hydroxide concentration from both water and the weak base to find the concentration of the hydronium ion?

Thank you

#### Borek

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##### Re: Question about pH with Water
« Reply #1 on: September 30, 2016, 02:43:05 AM »
Not seeing the original problem it is hard to comment on, but yes - sometimes you have to take products of water autodissociation into account. It is not just adding them to what you calculated though, as the new concentration would shift the equilibrium a bit.
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#### galpinj

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##### Re: Question about pH with Water
« Reply #2 on: October 01, 2016, 11:53:31 AM »
Not seeing the original problem it is hard to comment on, but yes - sometimes you have to take products of water autodissociation into account. It is not just adding them to what you calculated though, as the new concentration would shift the equilibrium a bit.

Hi Borek,

The original problem was as follows:
100mL of 0.1M HF (Ka=6.8x10-4) was titrated with 100mL of 0.1M NaOH

The strong acid will completely neutralize HF, leaving us with 0.01 mol F- (0.01/.2L = 0.05M F-).
KaxKb = Kw, therefore Kb=1.47x10-11

[OH-] = square root of (1.47x10-11)(0.05)
[OH-] = 8.6x10-7M

How would we include the concentration of OH- from water into an equation like this? I assumed we would just add them...

Thank you

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