[thuanthuan]

Hi,
I have a question concerning the equlibrium of H+ and (OH)-

Given a 1.0 liter of water. Dissolve 1.0*10^(-8) mol NaOH into it. Assume that the volume of the solution remain unchanged... temperature 25`C
Calculating the pH, someone calculate [OH-] ( = 1.0*10^(-8) M) and calculate [H+] = 1.0*10^(-14) / [OH-]  =  1.0*10^(-6) M. Then pH = 6.

But I wonder if it's right. We know that:  H2O  <->   H+   +    (OH)-
Before dissolving, in 1 liter of water, there are 1.0*10^(-7) mol H+ and 1.0*10^(-7) mol OH-. The way of calculating above doesn't take into consideration this amount of ions. Furthermore, the concentration of [OH-] in water before is ten times greater than that of the added amount. So the result of that person is wrong, isn't it ? If we add 0.1mol NaOH, then that 1.0*10^(-7) mol OH- is neglectible, but this time ...

I think that solution was wrong. So I choose my complicated way to calculate pH, but the answer is almost the same as the "wrong" solution. I don't know why ? The number 1.0*10^(-7) mol OH- here is very much relative to (1.0*10^(-8) and it must make a big difference. But why the pHs are the same ?

[Sam (NG)]

Having had a look at the problem i would have to come to the same conclusion that you did, that you have to use the OH ion concentration from the NaOH additive to the OH conc in pure water because the concentration is so negligible.

Think if you will, about the mass of NaOH that you are adding to water here: 400 ng in 1 litre of water. It's sweet FA.

You would theoretically still expect the pH to be above 7, and i got 7.041.

[AWK]

This is a problem on the common ion effect. It can be solved through a quadratic equation based on ionic product of water Kw.
x(10^-8+x)=10^-14
where x is a concentration of H⁺ and OH^- from a dissociation of water in the presence of NaOH.
This give pH=7.022

[Sam (NG)]

This is a problem on the common ion effect. It can be solved through a quadratic equation based on ionic product of water Kw.
x(10^-8+x)=10^-14
where x is a concentration of H⁺ and OH^- from a dissociation of water in the presence of NaOH.
This give pH=7.022

You learn something new everyday :S/.

[Borek]

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

[thuanthuan]

Yes, I also got 7.0217. Thank you guys very much.

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

Thanks for the link!