[galpinj]

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]

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.

[galpinj]

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

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=pH-acid-base-solution